Chapter 7. Hypothesis Testing with One Sample

202 Chapter 7: Hypothesis Testing with One Sample12. Let σ be the population standard deviation of women biologists’ salaries. The claim is that this standarddeviation is greater than 3,000, or σ >3,000. If this is true then σ ≤3,000 must be false. Since σ ≤3,000includes equality, we let H 0 be σ =3,000 and we let H 1 be σ >3,000.In Exercises 13-20,find the critical z values. In each case, assume that the normal distribution applies.13. In a two-tailed test, the critical values are ±z α /2. Since α=0.05, α/2 = 0.025. The critical values are then±z α /2=±z .025=±1.96.14. In a two-tailed test, the critical values are ±z α /2. Since α=0.01, α/2 = 0.005. The critical values are then±z α /2=±z .005=±2.575.15. In a right-tailed test, the critical value is z α. Since α=0.01, the critical value is z α= z .01= 2.33.16. In a left-tailed test, the critical value is −z α. Since α=0.05, the critical value is −z α=−z .01=−1.645.17. Since H 1 is p≠0.17, this is a two-tailed test. In a two-tailed test, the critical values are ±z α /2. Since α=0.10, α/2= 0.05. The critical values are then ±z α /2=±z .05=±1.645.18. Since H 1 is p>0.18, this is a right-tailed test. In a right-tailed test, the critical value is z α. Since α=0.10, thecritical value is z α= z .10= 1.28.19. Since H 1 is p

Chapter 7: Hypothesis Testing with One Sample 20324. H 0 : p = 0.103, which makes q = 0.897, n = 800, and p ˆ = 0.12. The test statistic is thenz = p ˆ − p 0.12 − 0.103= = 0.017pq 0.103⋅ 0.897 0.0107 = 1.582n 800In Exercises 25-32, use the given information to find the P-value. (Hint: See Figure 7-6.)25. It is a right tailed test, so the P-value is the area to the right of the test statistic, z = 0.55. Using the methods of5-2, the P-value is 1-0.7088 = 0.2912.26. It is a left tailed test, so the P-value is the area to the left of the test statistic,z = -1.72. Using the methods of 5-2, the P-value is 0.0427.27. It is a two-tailed test, and the test statistic z = 1.95, is to the right of center so the p-value is the twice the areato the right of the test statistic. Using the methods of 5-2, the P-value is 2×(1–0.9744) = 0.0512.28. It is a two-tailed test, and the test statistic z = -1.63, is to the left of center so the p-value is the twice the area tothe left of the test statistic. Using the methods of 5-2, the P-value is 2×(0.0516) = 0.1032.29. Since H 1 : p > 0.29, it is a right tailed test, so the P-value is the area to the right of the test statistic, z = 1.97.Using the methods of 5-2, the P-value is 1-0.9756 = 0.0244.30. Since H 1 : p ≠ 0.30, it is a two-tailed test, and the test statistic z = 2.44, is to the right of center so the p-value isthe twice the area to the right of the test statistic. Using the methods of 5-2, the P-value is 2×(1–0.9927) =0.0146.31. Since H 1 : p ≠ 0.31, it is a two-tailed test, and the test statistic z = 0.77, is to the right of center so the p-value isthe twice the area to the right of the test statistic. Using the methods of 5-2, the P-value is 2×(1–0.7794) =0.4412.32. Since H 1 : p < 0.32, it is a left tailed test, so the p-value is the area to the left of the test statistic, z = -1.90.Using the methods of 5-2, the P-value is 0.0287.In Exercises 33-36, state the final conclusion in simple, non-technical terms. Be sure to address the originalclaim. (Hint: See Figure 7-7)33. The original claim did not include equality, H 0 was rejected, so the conclusion is:The sample data support the claim that the proportion of men who are married is greater than 0.5.34. The original claim did not include equality, H 0 was rejected, so the conclusion is:The sample data support the claim that the proportion of college graduates who smoke is less than 0.2735. The original claim did not include equality, H 0 was not rejected, so the conclusion is:There is no sufficient sample evidence to support the claim that the proportion of fatal commercial aviationcrashes is different from 0.038.36. The original claim did include equality, H 0 was rejected, so the conclusion is:There is sufficient evidence to warrant the rejection of the claim that the proportion of M&Ms that are blue isequal to 0.10.In exercises 37-40, identify the type I error and the type II error that correspond to the given hypothesis.37. Type I: Conclude there is sufficient evidence to support that the proportion of married women is greater than0.5 when in actuality p = 0.5Type II: Fail to reject that the proportion of married women is 0.5 when in actuality p > 0.5.

204 Chapter 7: Hypothesis Testing with One Sample38. Type I: Conclude there is sufficient evidence to support that the proportion of college graduate who smoke isless than 0.27 when in actuality p = 0.27Type II: Fail to reject that the proportion of college graduate who smoke is 0.27 when in actuality p < 0.27.39. Type I: Conclude there is sufficient evidence to support that the proportion of fatal commercial aviationcrashes is different than 0.038 when in actuality p = 0.038Type II: Fail to reject that the proportion of fatal commercial aviation crashes is 0.038 when in actuality p ≠0.038.40. Type I: Conclude there is sufficient evidence to support that the proportion of M&Ms that are blue is differentthan 0.10 when in actuality p = 0.10Type II: Fail to reject that the proportion of M&Ms that are blue is 0.10 when in actuality p ≠ 0.10.41. The alternate hypothesis is H 1 p > 0.50, which makes this a right tailed test. The sample proportion isp ˆ = 0.27, which is less than 0.50. The test statistic is then negative. The critical region is to the right of z = 0,and so the test statistic cannot fall in the critical region, and so there is no chance to reject H 0 .42. Not necessarily. Since the significance 0.05 is larger than the significance 0.01, the critical values for α = 0.05are closer to z = 0 than those for α = 0.01. It is possible that the critical value for the hypothesis test fellbetween these critical values, and so was rejected at the 5% level and not at the 1% level. It is also possiblethat the test statistic was within the critical region for α = 0.01 also, and so it may be that null hypothesis wasrejected at both significance levels.43. We follow the steps outlined in the problem:a. Step 1The significance level is 5%, this is a left tailed test, so the critical value is −z 0.05=−1.645.Step 2Solving for p ˆ , and substituting the values p=0.5, n = 1998, and q = 0.5, we findpq0.5⋅ 0.5p ˆ = p − z .05= 0.5−1.645n 1998 = 0.48Step 3We now find the area to the left of p ˆ , using the z-score based on a population proportion p = 0.45.z = p ˆ − p 0.48 − 0.45= = 2.70pq 0.45⋅ 0.55n 1998The area to the left of 2.70 is 0.9965, which is the power of the test. In this hypothesis test, it would be verylikely that we reject H 0 when the true population proportion is p = 0.45.b. β = (1 – 0.9965) = 0.003544. We will solve the problem by first solving the z score for p ˆ using the critical value for α = 0.05 and the nullhypothesis value for p = 0.04. Doing so, we see thatpq0.4 × 0.6p ˆ = p + z = 0.4 −1.645 = 0.4 − 0.80588 .n nnFor a power of 0.80, we need to find the number with 0.80 area to its left under the standard normal curve.Table A-2 gives the value z = -0.84.Now, using the alternate value for p = 0.3, we again solve for p ˆ .pq0.3× 0.7p ˆ = p + z = 0.3− 0.84 = 0.3− 0.38494 .n nnSetting the left hand sides equal to one another, we get the following equation.0.3− 0.38494 = 0.4 − 0.80589 .nn

Chapter 7: Hypothesis Testing with One Sample 205Multiplying each side by n , and then solving for n, we find0.3 n − 0.38494 = 0.4 n − 0.80588−0.1 n =−0.42094n = 17.7So the required sample size is n = 18.7-3 Testing a Claim About a Proportion1. Mendel’s Hybridization Experimentsa. The sample proportion is p ˆ = 0.2494 , the claim is that the proportion is 0.25, so p = 0.25. The sample sizeis n = 8023. The requirements are met, since np = 8023× 0.25 = 2005.75. and nq = 8023× 0.75 = 6017.25.The test statistic isz = p ˆ − p 0.2494 − 0.25= =−0.124pq 0.25× 0.75n 8023b. The significance level is 0.05, and the claim is that the proportion is 0.25, making this a two tailed test. Thecritical values are ±z α /2=±z .025=±1.96.c. Since this is a two tailed test with test statistic to the left of center, the P-value is twice the area to the left ofthe test statistic. This area is found using Table A-2, and is 0.4522. So, P − value = 2 × 0.4522 = 0.9044 .d. Since the P-value is larger than the significance level, we would fail to reject H 0 . The conclusion is: Thereis not sufficient evidence to warrant the rejection of the claim that the proportion of green-flowered peas is0.25.e. No. A hypothesis test can only demonstrate sufficient evidence that a parameter is not a specified value, notthat the parameter is some value.2. Survey of Drinkinga. The sample proportion is p ˆ = 0.62, the claim is that the proportion is 0.50, so p = 0.50. The sample size is n= 1087. The requirements are met, since np = 1087 × 0.50 = 543.5. and nq = 1087 × 0.50 = 543.5. The teststatistic isz = p ˆ − p 0.62 − 0.50= = 7.913pq 0.50 × 0.50n 1087b. The significance level is 0.05, and the claim is that the proportion greater than 0.50, making this a righttailed test. The critical value is z α= z .05= 1.645.c. Since this is a right-tailed test with test statistic to the right of center, the P-value is the area to the right ofthe test statistic. This area is found using Table A-2, and is 0.0001. So, P − value = 0.0001.d. Since the P-value is smaller than the significance level, we would reject H 0 . The conclusion is: The sampledata support the claim that the majority of adults use alcoholic beverages.e. No. Small sample sizes in similar tests could have results that do not cause us to reject H 0 .In Exercises 3-6, assume that a method of gender selection is being tested with couples wanting to have babygirls. Identify the null hypothesis, alternate hypothesis, test statistic, P-value or critical value(s), conclusion aboutthe null hypothesis, and final conclusion that addresses the original claim. Use the P-value method unless yourinstructor specifies otherwise.3. The claim is that the proportion of girls born using the method is greater than 0.50, so this is a right tailed test.The sample size is n = 100 and the number of births that were girls was x = 65, so p ˆ = 65/100 = 0.65. Thesignificance level is 0.01.H 0 : p = 0.50H 1 : p > 0.50

Chapter 7: Hypothesis Testing with One Sample 207In a right-tailed test at the 0.01 significance level, the critical value is z α= z .01= 2.33.Since this is a right tailed test, the P-value is the area to the right of the test statistic. Using Table A-2, we findthat the P − value = 1− 0.9943 = 0.0057.We reject the null hypothesis.The sample data support the claim that the proportion of girls born when using this method of gender selectionis greater than 0.5.In Exercises 7-18, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-valueor critical value(s), conclusion about the null hypothesis, and final conclusion that addresses the original claim.Use the P-value method unless you instructor specifies otherwise.7. Cloning SurveyFirst, we check the requirements. The sample appears to be a simple random sample The conditions for abinomial are satisfied. The sample size is 1012 with the claim that less than 10% of adults say human cloningshould be allowed, making p = 0.10, so np = 1012 × 0.10 = 101.2 and nq = 1012 × 0.90 = 910.8. Therequirements are satisfied.The claim is that less than 10% of all adults say that human cloning should be allowed, so this is a left-tailedtest. The sample proportion is p ˆ = 0.09.H 0 : p = 0.10H 1 : p < 0.10The test statistic is z = p ˆ − p =pqn0.09 − 0.100.10 × 0.901012=−1.060In a left-tailed test at the 0.05 significance level, the critical value is −z α=−z .05=−1.645.Since this is a left tailed test, the P-value is the area to the left of the test statistic. Using Table A-2, we findthat the P − value = 0.1446.We fail to reject the null hypothesis.There is not sufficient sample evidence to support the claim that less than 10% of all adults say that cloning ofhumans should be allowed. A newspaper could run the headline, but the headline is unsupported by the sampleevidence.8. Testing Lipitor for Cholesterol ReductionFirst, we check the requirements. The sample does not appear to be a simple random sample, but the subjectsmay comprise a random sample that is representative of the population. The conditions for a binomial aresatisfied. The sample size is 863 with the claim that the percentage of treated patients with flu symptoms isgreater than the 1.9% rate for patients not given the treatments, making p = 0.019, sonp = 863× 0.019 = 16.397 and nq = 863× 0.981 = 846.603. The requirements are satisfied.The claim is that more than 1.9% of all those receiving Lipitor treatments have flu symptoms, so this is a righttailedtest. In the sample, 19 had flu symptoms. The sample proportion is p ˆ = x / n = 19 /863 = 0.022.H 0 : p = 0.019H 1 : p > 0.019The test statistic is z = p ˆ − p 0.022 − 0.019= = 0.646pq 0.019 × 0.981n 863In a right-tailed test at the 0.01 significance level, the critical value is z α= z .01= 2.33.Since this is a right tailed test, the P-value is the area to the right of the test statistic. Using Table A-2, we findthat the P − value = (1 − 0.7422) = 0.2578.We fail to reject the null hypothesis.There is not sufficient sample evidence to support the claim that more than 1.9% of all those receiving Lipitortreatments have flu symptoms. It does not seem that flu symptoms are an adverse reaction to the treatment.

208 Chapter 7: Hypothesis Testing with One Sample9. Cell Phones and CancerFirst, we check the requirements. The sample does not appear to be a simple random sample, but the subjectsmay comprise a random sample that is representative of the population. The conditions for a binomial aresatisfied. The sample size is 420,095 with the claim that the rate at which cell phone users develop cancer ofthe brain or nervous system is different than 0.0340%, making p = 0.00034, sonp = 420,095× 0.000340 = 142.8323 and nq = 420,095× 0.999660 = 419,952.168. The requirements aresatisfied.The claim is that the rate at which cell phone users develop cancer of the brain or nervous system is differentthan 0.0340%, so this is a two-tailed test. In the sample, 135 developed cancer of the brain or nervous system.The sample proportion is p ˆ = x / n = 135/420,095 = 0.000321.H 0 : p = 0.000340H 1 : p 0.000340≠The test statistic is z = p ˆ − p 0.000321− 0.000340= =−0.668pq 0.000340× 0.999660n420,095In a two-tailed test at the 0.005 significance level, the critical values are ±z α /2=±z .0025=±2.81.Since this is a two-tailed test and the test statistic is to the left of center, the P-value is twice the area to the leftof the test statistic. Using Table A-2, we find that the P − value = 2 × (0.2514) = 0.5028.We fail to reject the null hypothesis.There is not sufficient sample evidence to support the claim that the rate of cancer of the brain or nervoussystem is different than the 0.0340% rate for those who do not use cell phones. Users of cell phones need notbe more concerned with cancer of the brain or nervous system than those who do not use cell phones.10. Drug Testing of Job ApplicantsFirst, we check the requirements. The sample is a simple random sample. The conditions for a binomial aresatisfied. The sample size is 1520 with the claim that the failure rate of job applicants for drugs is less than5.8%, making p = 0.058, so np = 1520 × 0.058 = 88.16 and nq = 1520 × 0.942 = 1431.84. The requirements aresatisfied.The claim is that the drug test failure rate of job applicants is less than 5.8%, so this is a left-tailed test. In thesample, 58 applicants failed the drug test. The sample proportion is p ˆ = x / n = 58 /1520 = 0.0382.H 0 : p = 0.058H 1 : p < 0.058The test statistic is z = p ˆ − p 0.0382 − 0.058= =−3.303pq 0.058 × 0.942n 1520In a left-tailed test at the 0.01 significance level, the critical value is −z α=−z .01=−2.33.Since this is a left tailed test, the P-value is the area to the left of the test statistic. Using Table A-2, we findthat the P − value = 0.0005.We reject the null hypothesis.The sample data support the claim that the drug test failure rate for job applicants is now lower than the 1990failure rate of 5.8%. It may suggest this, but there is also the possibility that job applicants, now familiar withthe drug tests and their limitations, are better at not being caught by those tests.11. Testing Effectiveness of Nicotine PatchesFirst, we check the requirements. The sample does not appear to be a simple random sample, but the subjectsmay comprise a random sample that is representative of the population. The conditions for a binomial aresatisfied. The sample size is 39 + 32 = 71 with the claim that the majority of smokers who try to quit withnicotine patch therapy are smoking one year after the treatment, making p = 0.50, so np = 71× 0.50 = 35.5 andnq = 71× 0.50 = 35.5. The requirements are satisfied.

Chapter 7: Hypothesis Testing with One Sample 209The claim is that the majority of smokers who try to quit with nicotine patch therapy are smoking one yearafter the treatment, so this is a right-tailed test. In the sample, 39 were smoking one year after treatment. Thesample proportion is p ˆ = x / n = 39 /71 = 0.549.H 0 : p = 0.50H 1 : p > 0.50The test statistic is z = p ˆ − p 0.549 − 0.50= = 0.826pq 0.50 × 0.50n 71In a right-tailed test at the 0.10 significance level, the critical value is z α= z .10= 1.28.Since this is a right tailed test, the P-value is the area to the right of the test statistic. Using Table A-2, we findthat the P − value = (1 − 0.7967) = 0.2033.We fail to reject the null hypothesis.There is not sufficient sample evidence to support the claim that the majority of smokers who try to quit withnicotine patch therapy are smoking one year after the treatment. Although a majority return to smoking oneyear after the treatment, this does not mean the treatment is ineffective. To be certain, the rate at which thetherapy patients return to smoking should be compared to others who have quit smoking.12. Smoking and College EducationFirst, we check the requirements. The sample is a random sample. The conditions for a binomial are satisfied.The sample size is 785 with the claim that the percentage of people with four years of college that smoke isless than 27%, making p = 0.27, so np = 785× 0.27 = 211.95 and nq = 785× 0.73 = 573.05. The requirementsare satisfied.The claim is that the percentage of people with four years of college that smoke is less than 27%, so this is aleft-tailed test. In the sample, 144 subjects with four years of college education smoked. The sample proportionis p ˆ = x / n = 144 /785 = 0.183.H 0 : p = 0.27H 1 : p < 0.27The test statistic is z = p ˆ − p =pqn0.183− 0.270.27 × 0.73785=−5.490In a left-tailed test at the 0.01 significance level, the critical value is −z α=−z .01=−2.33.Since this is a left tailed test, the P-value is the area to the left of the test statistic. Using Table A-2, we findthat the P − value = 0.0001.We reject the null hypothesis.The sample data support the claim that the percentage of people with four years of college that smoke is lessthan 27%. One reason that they may smoke at a lower rate is that they are better educated on the hazards anddangers.13. Drug Testing for Adverse ReactionsFirst, we check the requirements. The sample does not appear to be a simple random sample, but the subjectsmay comprise a random sample that is representative of the population. The conditions for a binomial aresatisfied. The sample size is 221 with the claim that the percentage of users who experience dizziness is lessthan 5%, making p = 0.05, so np = 221× 0.05 = 11.05 and nq = 221× 0.95 = 209.95. The requirements aresatisfied.The claim is that the percentage of users who experience dizziness is less than 5%, so this is a left-tailed test.As stated in the exercise, p ˆ = .032.H 0 : p = 0.05H 1 : p < 005The test statistic is z = p ˆ − p =pqn0.032 − 0.05=−1.2280.05× 0.95221

210 Chapter 7: Hypothesis Testing with One SampleIn a left-tailed test at the 0.01 significance level, the critical value is −z α=−z .01=−2.33.Since this is a left tailed test, the P-value is the area to the left of the test statistic. Using Table A-2, we findthat the P − value = 0.1093.We fail to reject the null hypothesis.There is not sufficient sample evidence to support the claim that the percentage of users of Ziac whoexperience dizziness is less than 5%.14. Aviation FatalitiesFirst, we check the requirements. The sample does not appear to be a simple random sample, but the incidentsmay comprise a random sample that is representative of the population. The conditions for a binomial aresatisfied. The sample size is 8411 with the claim that the percentage of crash landings that result in a fatality isgreater than 5%, making p = 0.05, so np = 8411× 0.05 = 420.55 and nq = 8411× 0.95 = 7990.45. Therequirements are satisfied.The claim is that pilots die in more than 5% of all crash landings, so this is a right-tailed test. In the sample,437 pilots died in crash landings. The sample proportion is p ˆ = x / n = 437 /8411 = 0.05196.H 0 : p = 0.05H 1 : p > 0.05The test statistic is z = p ˆ − p 0.05196 − 0.05= = 0.825pq 0.05× 0.95n 8411In a right-tailed test at the 0.06 significance level, the critical value is z α= z .06= 1.555.Since this is a right tailed test, the P-value is the area to the right of the test statistic. Using Table A-2, we findthat the P − value = (1 − 0.7967) = 0.2033.We fail to reject the null hypothesis.There is not sufficient sample evidence to support the claim that pilots die in more than 5% of all crashlandings. However, we cannot conclude that the fatality rate does not exceed 5%, since all we have seen is thatthere is not enough to show the rate exceeds 5%. It still might. We simply don’t have enough evidence in thissample.15. Air-Bag EffectivenessFirst, we check the requirements. The sample does not appear to be a simple random sample, but the incidentsmay comprise a random sample that is representative of the population. The conditions for a binomial aresatisfied. The sample size is 821 with the claim that the air-bag hospitalization rate is lower than 7.8%, makingp = 0.078, so np = 821× 0.078 = 64.038 and nq = 821× 0.922 = 756.962. The requirements are satisfied.The claim is that the air-bag hospitalization rate is lower than 7.8%, so this is a left-tailed test. The sampleproportion is p ˆ = x / n = 46 /821 = 0.0560.H 0 : p = 0.078H 1 : p < 0.078The test statistic is z = p ˆ − p =pqn0.0560 − 0.0780.078 × 0.922821=−2.351In a left-tailed test at the 0.01 significance level, the critical value is −z α=−z .01=−2.33.Since this is a left tailed test, the P-value is the area to the left of the test statistic. Using Table A-2, we findthat the P − value = 0.0094.We reject the null hypothesis.The sample data support the claim that the air-bag hospitalization rate is lower than the 7.8% rate for crashesof midsized cars equipped with automatic safety belts.

Chapter 7: Hypothesis Testing with One Sample 21116. Birds and Aircraft CrashesFirst, we check the requirements. The sample does not appear to be a simple random sample, but the subjectsmay comprise a random sample that is representative of the population. The conditions for a binomial aresatisfied. The sample size is 74 with the claim that 10% of aborted takeoffs leading to aircraft going off therunway are due to bird strikes, making p = 0.10, so np = 74 × 0.10 = 7.4 and nq = 74 × 0.90 = 66.6. Therequirements are satisfied.The claim is that 10% of aborted takeoffs leading to aircraft going off the runway are due to bird strikes, sothis is a two-tailed test. In the sample, 5 aborted takeoffs that led to aircraft going off the runway were due tobird strikes. The sample proportion is p ˆ = x / n = 5/74 = 0.0676.H 0 : p = 0.10H 1 : p 0.10 ≠The test statistic is z = p ˆ − p 0.0676 − 0.10= =−0.929pq 0.10 × 0.90n 74In a two-tailed test at the 0.05 significance level, the critical values are ±z α /2=±z .025=±1.96.Since this is a two-tailed test and the test statistic is to the left of center, the P-value is twice the area to the leftof the test statistic. Using Table A-2, we find that the P − value = 2 × 0.1762 = 0.3524 .We fail to reject the null hypothesis.There is not sufficient sample evidence to warrant rejection of the claim that 10% of aborted takeoffs that leadto aircraft going off the runway are due to bird strikes.17. Females Underrepresented in TextbooksFirst, we check the requirements. The sample does not appear to be a simple random sample, but theillustrations may comprise a random sample that is representative of the population. The conditions for abinomial are satisfied. The sample size is 142 + 554 = 696 with the claim that the majority of nonreproductiveillustrations are of males, making p = 0.50, so np = 696 × 0.50 = 348 and nq = 696 × 0.50 = 348. Therequirements are satisfied.The claim is that the majority of nonreproductive illustrations are of males, so this is a right-tailed test. In thesample, 554 nonreproductive illustrations depicted males. The sample proportion isp ˆ = x / n = 554 /696 = 0.796.H 0 : p = 0.50H 1 : p > 0.50The test statistic is z = p ˆ − p 0.796 − 0.50= = 15.618pq 0.50 × 0.50n 696In a right-tailed test at the 0.05 significance level, the critical value is z α= z .05= 1.645.Since this is a right tailed test, the P-value is the area to the right of the test statistic. Using Table A-2, we findthat the P − value = (1 − 0.9999) = 0.0001.We reject the null hypothesis.There is sufficient sample evidence to support the claim that the majority of nonreproductive illustrations areof males.

212 Chapter 7: Hypothesis Testing with One Sample18. Fruit FliesFirst, we check the requirements. The sample does not appear to be a simple random sample, but the fruit fliesmay comprise a random sample that is representative of the population. The conditions for a binomial aresatisfied. The sample size is 80 with the claim that the percentage of offspring with normal wings is differentthan 75%, making p = 0.75, so np = 80 × 0.75 = 60 and nq = 80 × 0.25 = 20. The requirements are satisfied.The claim is that the percentage of offspring with normal wings is different than 75%, so this is a two-tailedtest. As stated, p ˆ = 0.85.H 0 : p = 0.75H 1 : p 0.75 ≠The test statistic is z = p ˆ − p 0.85− 0.75= = 2.066pq 0.75× 0.25n 80In a two-tailed test at the 0.01 significance level, the critical values are ±z α /2=±z .005=±2.575.Since this is a two-tailed test and the test statistic is to the right of center, the P-value is twice the area to theright of the test statistic. Using Table A-2, we find that the P − value = 2 × (1 − 0.9808) = 0.0384 .We fail to reject the null hypothesis.There is not sufficient sample evidence to warrant rejection of the claim that the percentage of offspring withnormal wings is different than 75%.19. Using Confidence Intervals to Test Hypothesisa. First, we check the requirements. The sample is a random sample. The conditions for a binomial aresatisfied. The sample size is 1000 with the claim that the proportion of zeros is 0.1, making p = 0.1, sonp = 1000 × 0.10 = 100 and nq = 1000 × 0.90 = 900. The requirements are satisfied.The claim is that the proportion of zeros is 0.1, so this is a two-tailed test. In the sample, 119 were zero.The sample proportion is p ˆ = x / n = 119 /1000 = 0.119.H 0 : p = 0.1H 1 : p 0.1 ≠The test statistic is z = p ˆ − p 0.119 − 0.1= = 2.003pq 0.1× 0.9n 1000In a two-tailed test at the 0.05 significance level, the critical values are ±z α /2=±z .025=±1.96.There is sufficient evidence to warrant the rejection of the claim that the proportion of last digits in weightsis 0.1.b. The preliminaries are the same as in part a.H 0 : p = 0.1H 1 : ≠ p 0.1The test statistic is z = p ˆ − ppqn0.119 − 0.1= = 2.0030.1× 0.91000Since this is a two-tailed test and the test statistic is to the right of center, the P-value is twice the area to theright of the test statistic. Using Table A-2, we find that the P − value = 2 × (1 − 0.9772) = 0.0456.There is sufficient evidence to warrant the rejection of the claim that the proportion of last digits in weightsis 0.1.c. We use the information from part a. On page 261, the table of confidence levels and critical valuesindicates that for a confidence level of 95%, the critical value is 1.96. The margin of error is given byp ˆ q ˆE = z α 2 n = (1.96) (0.119)(0.881)= (1.96)(0.010) = 0.020.1000So the confidence interval is( p ˆ − E, p ˆ + E) = (0.119 − 0.020,0.119 + 0.020) = (0.099,0.139)The confidence interval seems to suggest that the proportion of zeros being 0.1 is not out of the question.d. They do not all lead to the same conclusion. The two hypothesis tests lead to a rejection of the nullhypothesis but the confidence interval does not support that result.

Chapter 7: Hypothesis Testing with One Sample 21320. Proving ClaimsThe claim seems to be that the rate of leukemia in children near power lines is less than or equal to c, which isto say no more likely. Writing it out symbolically, H 0 : p = c and H 1 : p > c. If we fail to reject, ourconclusion will essentially be that there isn’t enough evidence to prove the proportion of children that livenear power lines and who get leukemia is greater than the proportion of children who do not live nearpower lines and get leukemia. If we reject, then we will say that there is sufficient evidence to warrantrejection of the claim that this proportion is no greater than c. In neither case is the claim supported.21. Power and Probability of Type II ErrorThe significance level is 5%, this is a right tailed test, so the critical value is z 0.05= 1.645.Solving for p ˆ , and substituting the values p=0.5, n = 50, and q = 0.5, we findpq0.5⋅ 0.5p ˆ = p + z .05= 0.5+1.645 = 0.616n 50We now find the area to the right of p ˆ , using the z-score based on a population proportion p = 0.45.z = p ˆ − p 0.616 − 0.45= = 2.359pq 0.45⋅ 0.55n 50The area to the right of 2.36 is 0.0091, which is the power of the test. In this hypothesis test, it would be veryunlikely that we reject H 0 when the true population proportion is p = 0.45. β = 1 – 0.0091 = .9909. There is ahigh probability we will commit a type II error, in that we would fail to reject an incorrect null hypothesis.7-4 Testing a Claim About a Mean: σ KnownIn Exercises 1-4, determine whether the given conditions justify using the methods of this section when testing aclaim about the population mean µ.Note: It is not known how data was collected, so we assume that the sample is a simple random sample.1. Since σ is known, and the original population is normally distributed, the conditions are met.2. Since σ is unknown, the conditions are not met.3. Since σ is unknown, the conditions are not met.4. Since σ is known, and the sample size (n = 47) is larger than 30, the conditions are met.In Exercises 5-8,find the test statistic, P-value or critical value(s), and state the final conclusion.5. The test statistic is z = x − µ 120 −118=σ / n 12 / 50 = 1.18The claim is that the mean IQ is greater than 118, making this a right tailed test. In a right-tailed test, at the0.05 significance level, the critical value is z α= z .05= 1.645.Since this is a right tailed test, the P-value is the area to the right of the test statistic. Using Table A-2, we findthat the P − value = 1− 0.8810 = 0.1190.There is not sufficient evidence to support the claim that the mean IQ of statistics professors is greater than118.

214 Chapter 7: Hypothesis Testing with One Sample6. The test statistic is z = x − µ 98.20 − 98.6=σ / n 0.62 / 106 =−6.642The claim is that the mean body temperature of healthy adults is less than 98.6°F, making this a left tailed test.In a left-tailed test at the 0.01 significance level, the critical value is −z α=−z .01=−2.33.Since this is a left tailed test, the P-value is the area to the left of the test statistic. Using Table A-2, we findthat the P − value = 0.0001.There is sufficient evidence to support the claim that mean body temperature of healthy adults is less than98.6°F.7. The test statistic is z = x − µ 5.25− 5.00=σ / n 2.50 / 80 = 0.894The claim is that the mean time between uses of a TV remote control by males during commercials is 5.00seconds, making this a two-tailed test. In a two-tailed test at the 0.01 significance level, the critical values are±z α /2=±z .005=±2.575.Since this is a two-tailed test and the test statistic is to the right of center, the P-value is twice the area to theright of the test statistic. Using Table A-2, we find that the P − value = 2 × (1 − 0.8133) = 0.3734.There is not sufficient evidence to warrant the rejection of the claim that the mean time between uses of a TVremote control by males during commercials is 5.00 seconds.8. The test statistic is z = x − µ 45,678 − 46,000= =−0.262σ / n 9900 / 65The claim is that the mean salary for college graduates who have taken a statistics course is equal to $46,000,making this a two-tailed test. In a two-tailed test at the 0.05 significance level, the critical values are±z α /2=±z .025=±1.96.Since this is a two-tailed test and the test statistic is to the left of center, the P-value is twice the area to the leftof the test statistic. Using Table A-2, we find that the P − value = 2 × (0.3974) = 0.7948.There is not sufficient evidence to warrant the rejection of the claim that the mean salary for college graduateswho have taken a statistics course is equal to $46,000In Exercises 9-12, test the given claim. Identify the null hypothesis, alternate hypothesis, test statistic, P-value orcritical value(s), conclusion about the null hypothesis, and final conclusion that addresses the original claim. Usethe P-value method unless your instructor specifies otherwise.9. Everglades TemperaturesFirst, we check the conditions. The sample size (n = 61) is greater than 30, and we are assuming a value of σ.However, the sample may not be a simple random sample. However, we will continue as if it were. The claimis that the mean temperature is greater than 30.0°C, so this is a right tailed test. The sample mean is 30.4°C andthe assumed population standard deviation is 1.7°C. The significance level is 0.05.H 0 : µ = 30.0H 1 : µ > 30.0The test statistic is z = x − µ 30.4 − 30.0=σ / n 1.7 / 61 = 1.838In a right-tailed test at the 0.05 significance level, the critical value is z α= z .05= 1.645.Since this is a right tailed test, the P-value is the area to the right of the test statistic. Using Table A-2, we findthat the P − value = 1− 0.9671 = 0.0329.We reject the null hypothesis.The sample data support the claim that the mean temperature is greater than 30.0°C.10. Weights of BearsFirst, we check the conditions. The sample size (n = 54) is greater than 30, and we are assuming a value of σ.However, the sample may not be a simple random sample. However, we will continue as if it were. The claimis that the mean weight of the bears is less than 200 lb, so this is a left tailed test. The sample mean is 182.9 lband the assumed population standard deviation is 121.8 lb. The significance level is 0.10.

Chapter 7: Hypothesis Testing with One Sample 215H 0 : µ = 200H 1 : µ < 200The test statistic is z = x − µ 182.9 − 200=σ / n 121.8 / 54 =−1.032In a left-tailed test at the 0.10 significance level, the critical value is −z α=−z .05=−1.28 .Since this is a left tailed test, the P-value is the area to the left of the test statistic. Using Table A-2, we findthat the P − value = 0.1515.We fail to reject the null hypothesis.There is not sufficient sample evidence to support the claim that the mean weight is less than 200 lb.11. Cotinine Levels of SmokersFirst, we check the conditions. The sample size (n = 40) is greater than 30, we are assuming a value of σ, andthe sample is a random sample. The claim is that the mean cotinine level of all smokers is 200, so this is a twotailedtest. The sample mean is 172.5 and the assumed population standard deviation is 119.5. The significancelevel is 0.01.H 0 : µ = 200H 1 : µ 200 ≠The test statistic is z = x − µ 172.5− 200=σ / n 119.5/ 40 =−1.455In a two-tailed test at the 0.01 significance level, the critical values are ±z α /2=±z .005=±2.575.Since this is a two-tailed test and the test statistic is to the left of center, the P-value is twice the area to the leftof the test statistic. Using Table A-2, we find that the P − value = 2 × (0.0735) = 0.1470.We fail to reject the null hypothesis.There is not sufficient evidence to warrant the rejection of the claim that the mean cotinine levels of smokers is200.0.12. Head CircumferencesFirst, we check the conditions. The sample size (n = 100) is greater than 30, we are assuming a value of σ, andthe sample is a random sample. The claim is that the mean head circumference for two month old babies is40.0 cm, so this is a two-tailed test. The sample mean is 40.6 cm and the assumed population standarddeviation is 1.6 cm. The significance level is 0.05.H 0 : µ = 40.0H 1 : µ ≠ 40.0The test statistic is z = x − µ 40.6 − 40.0=σ / n 1.6 / 100 = 3.75In a two-tailed test at the 0.05 significance level, the critical values are ±z α /2=±z .025=±1.96.Since this is a two-tailed test and the test statistic is to the right of center, the P-value is twice the area to theright of the test statistic. Using Table A-2, we find that the P − value = 2 × (1 − 0.9999) = 0.0002.We reject the null hypothesis.There is sufficient evidence to warrant the rejection of the claim that the mean head circumference for twomonth old babies is 40.0 cm.

216 Chapter 7: Hypothesis Testing with One Sample13.Testing the Assumed σa. That the population standard deviation is actually known.b. The significance level of the test was 0.05, it was a two tailed test, so any test statistic at least as small as−z .025=−1.96 would result in the reject of the null hypothesis. Solving the test statistic formula for σ yieldsσ = x − µz / n .Substituting n = 106, x = 98.2, µ = 98.6, and z = -1.96, we find the largest value for σ that results in arejection of the null hypothesis.σ = x − µ 98.2 − 98.6=z / n −1.96 / 106 = 2.1011c. Yes, there is a reasonable chance that the population standard deviation is greater than 0.62. Since it wasbased on a sample, we know that it is highly unlikely that the true value of σ is 0.62. The assumption that σis 0.62 is not realistic, but that value may be very close.14. Power and probability of Type II ErrorThe significance level is 0.05, this is a two-tailed test, so the critical values are ±z 0.025=±1.96 .Solving for x , and substituting the values µ=98.6, n = 106, and σ = 0.62, we find two values for xσ0.62x = µ ± z .025= 98.6 ±1.96n 106 = 98.718,98.482We now find the area to the right of each value of x , using the z-score based on a population mean µ = 98.4.z = x − µ 98.482 − 98.4x − µ 98.718 − 98.4= = 1.362 and z = =σ / n 0.62 / 106 σ / n 0.62 / 106 = 5.281For the first value, the power of the test is the area to the left of 1.36, which is 0.9131. For the second, thepower of the test is the area to the right of 5.281, which is 0.9999.b. The value of β = (1 – 0.9131) = 0.0869 and is also β = (1 – 0.9999) = 0.0001.15. Power of TestThe significance is 0.05 and this is a left-tailed test, so the critical value is −z 0.05=−1.645. Solving for x , andsubstituting the values µ=98.6, n = 106, and σ = 0.62, we find the value for xσ0.62x = µ − z .05= 98.6 −1.645n 106 = 98.501.Since the power is to be 0.8, this means that,under the assumption that the mean is really the alternate value,the z-score for 98.501 has area to its left equal to 0.8. Using table A-2 we find this value, 0.84. Solving the teststatistic for µ and substituting in the values gives the followingσµ = x + z .2n= 98.501+ 0.840.62106 = 98.55.7-5 Testing a Claim About a Mean: σ Not KnownIn Exercises 1-4, determine whether the hypothesis test involves a sampling distributions of means that is anormal distribution, Student t distribution, or neither.Note: It is not known how data was collected, so we assume that the sample is a simple random sample.1. The data appear to come from a normally distributed population, and σ is unknown, so this involves theStudent t distribution.2. The data appear to come from a distribution far from normal, the sample size, n = 25, is less than 30, and σ isunknown, so this involves neither.3. The data appear to come from a normally distributed population, and σ is known, so this involves the normaldistribution.

Chapter 7: Hypothesis Testing with One Sample 2174. The data appear to come from a distribution that is not normal, the sample size, n = 150, is larger than 30, andσ is unknown, so this involves the Student t distribution..In Exercises 5-8, use the given information to find a range of numbers for the P-value..5. In the row for 11 degrees of freedom, the test statistic falls between 3.106 and 2.718, so the P-value is between0.005 and 0.01.6. In the row for 11 degrees of freedom, the absolute value of the test statistic falls to the right of 1.363 so the P-value is more than 0.10.7. In the row for 15 degrees of freedom, the test statistic falls to the left of 2.947, so the P-value is less than 0.01.8. In the row for 8 degrees of freedom, the absolute value of the test statistic falls between 1.860 and 1.397, sothe P-value is between 0.10 and 0.20.In Exercises 9-12, assume that a simple random sample has been selected from a normally distributedpopulation. test the given claim. Find the test statistic, P-value or critical value(s), and state the final conclusion.9. The claim is that the mean IQ score of statistics professors is greater than 118, so this is a right-tailed test. Thesample size is n = 20, the sample mean is x = 120, and the sample standard deviation is s = 12. Thesignificance level is 0.05.The test statistic is t = x − µ 120 −118=s / n 12 / 20 = 0.745In a right-tailed test at the 0.05 significance level with 19 degrees of freedom, the critical value ist α= t .05= 1.729.Since the test statistic is smaller than 1.328, the P-value is greater than 0.10.We fail to reject the null hypothesis.There is not sufficient sample evidence to support the claim that the mean IQ of statistics professors is greaterthan 118.10. The claim is that the mean body temperature of healthy adults is less than 98.6°F, so this is a left-tailed test.The sample size is n = 35, the sample mean is x = 98.20, and the sample standard deviation is s = 0.62. Thesignificance level is 0.01.The test statistic is t = x − µ 98.20 − 98.60= =−3.817s / n 0.62 / 35In a left-tailed test at the 0.01 significance level with 34 degrees of freedom, the critical value is−t α=−t .01=−2.441.Since the absolute value of the test statistic is larger than 2.728, the P-value is less than 0.005.We reject the null hypothesis.The sample data support the claim that the mean body temperature of healthy adults is less than 98.6°F.11. The claim is that the mean time between uses of a TV remote control by males during commercials equals 5.00sec, so this is a two-tailed test. The sample size is n = 81, the sample mean is x = 5.25, and the samplestandard deviation is s = 2.50. The significance level is 0.01.The test statistic is t = x − µ 5.25− 5.00=s / n 2.50 / 81 = 0.90In a two-tailed test at the 0.01 significance level with 80 degrees of freedom, the critical values are±t α /2=±t .005=±2.639.Since the test statistic is smaller than 1.292, the P-value is greater than 0.20.We fail to reject the null hypothesis.There is not sufficient evidence to warrant the rejection of the claim that the mean time between uses of a TVremote control by males during commercials equals 5.00 sec.

218 Chapter 7: Hypothesis Testing with One Sample12. The claim is that the mean starting salary for college graduates who have taken a statistics course is equal to$46,000, so this is a two-tailed test. The sample size is n = 27, the sample mean is x = 45,678, and the samplestandard deviation is s = 9,900. The significance level is 0.05.The test statistic is t = x − µ 45,678 − 46,000= =−0.169s / n 9,900 / 27In a two-tailed test at the 0.05 significance level with 26 degrees of freedom, the critical values are±t α /2=±t .025=±2.056.Since the absolute value of the test statistic is smaller than 1.315, the P-value is greater than 0.20.We fail to reject the null hypothesis.There is not sufficient evidence to warrant the rejection of the claim that the mean salary for college graduateswho have taken a statistics course is equal to $46,000.In Exercises 13-24, assume that a simple random sample has been selected from a normally distributedpopulation and test the given claim. Unless specified by your instructor, use either the traditional method orthe P-value method for testing hypotheses.13. Effect of Vitamin Supplement on Birth WeightThe claim is that the mean birth weight for all male babies of mothers given vitamins is 3.39 kg, so this is atwo-tailed test. The P-value is provided by the output, and is 0.103, and we would fail to reject the nullhypothesis. There is not sufficient evidence to warrant the rejection of the claim that the mean birth weight forall male babies of mothers given vitamins is 3.39 kg. In other words, there isn’t enough evidence to say thatthe vitamin supplements have an effect on birth weight.14. Pulse RatesThe claim is that the mean pulse rate for statistics students is greater than 60 beats per minute. The P-value forthe claim is displayed with the calculator output, and is 0.00000197. We would reject the null hypothesis. Thesample data support the claim that the mean pulse rate for statistics students is greater than 60 beats perminute.15. Heights of ParentsThe claim is that women tend to marry men who are taller than themselves, or that the mean differencebetween husbands and wives is greater than zero, so this is a right-tailed test. The sample size is n = 20 makingthe degrees of freedom df = 19, the sample mean is x = 4.4 , and the sample standard deviation is s = 4.2. Thesignificance level is 0.01.H 0 : µ = 0H 1 : µ > 0The test statistic is t = x − µs / n = 4.4 − 04.2 / 20 = 4.685In a right-tailed test at the 0.01 significance level with df = 19, the critical value is t α= t .01= 2.539.In the row for 19 degrees of freedom, the test statistic falls to the left of 2.861 so the P-value is less than 0.005.We reject the null hypothesis.The sample data support the claim that women tend to marry men who are taller than themselves.16. Monitoring Lead in AirThe claim is that the amount of lead in the air in Building 5 of the World Trade Center following the collapseof the two building is greater than 1.5 µg/m 3 , so this is a right-tailed test. The sample size is n = 6 making thedegrees of freedom df = 5. The significance level is 0.01. Calculating x from the sample data givesx = Σx /n = 9.23/6 = 1.538Calculating the sample standard deviation from the data givess = nΣ(x 2 ) − (Σx) 226 ⋅ 32.520 − (9.23)= = 1.914n(n −1)6 ⋅ 5We move on to the test.H 0 : µ = 1.5H 1 : µ > 1.5

Chapter 7: Hypothesis Testing with One Sample 219The test statistic is t = x − µ 1.538 −1.5=s / n 1.914 / 5 = 0.044In a right-tailed test at the 0.01 significance level with df = 5, the critical value is t α= t .01= 3.365.In the row for 5 degrees of freedom, the test statistic lies above 1.476, so the P-value is greater than 0.10.We fail to reject the null hypothesis.There is not sufficient sample evidence to support the claim that the amount of lead in the air in Building 5 ofthe World Trade Center following the collapse of the two building is greater than 1.5 µg/m 3 .The assumption that the data came from a normally distributed population seems unrealistic, as there is anextreme outlier in the data set.17. Sugar in CerealThe claim is that the mean sugar content for all cereals is les than 0.3g, so this is a left-tailed test. The samplesize is n = 16 making the degrees of freedom df = 15, the sample mean is x = 0.295, and the sample standarddeviation is s = 0.168. The significance level is 0.05.H 0 : µ = 0.3H 1 : µ < 0.3The test statistic is t = x − µ 0.295− 0.3=s / n 0.168 / 16 =−0.119In a left-tailed test at the 0.05 significance level with df = 15, the critical value is −t α=−t .05=−1.753.In the row for 15 degrees of freedom, the absolute value for the test statistic falls to the right of 1.341 so the P-value is greater than 0.10.We fail to reject the null hypothesis.There is not sufficient sample evidence to support the claim that the mean sugar content for all cereals is lessthan 0.3g.18. Treating Chronic Fatigue SyndromeThe claim is that the mean score on the fatigue scale is positive, or greater than zero, so this is a right-tailedtest. The sample size is n = 21 making the degrees of freedom df = 20. The significance level is 0.01.Calculating x from the sample data givesx = Σx /n = 84 /21 = 4.0Calculating the sample standard deviation from the data givess = nΣ(x 2 ) − (Σx) 2221⋅ 430 − (84)= = 2.168n(n −1)21⋅ 20We move on to the test.H 0 : µ = 0H 1 : µ > 0The test statistic is t = x − µs / n = 4 − 02.168 / 21 = 8.455In a right-tailed test at the 0.01 significance level with df = 20, the critical value is t α= t .01= 2.528.In the row for 20 degrees of freedom, the test statistic lies to the left of 2.845, so the P-value is less than 0.005.We reject the null hypothesis.The sample data support the claim that the mean score on the fatigue scale is positive. The treatments appear tobe effective.

220 Chapter 7: Hypothesis Testing with One Sample19. Olympic WinnersThe claim is that the mean winning time in the Olympic 100 m dash is less than 10.5 sec, so this is a left-tailedtest. The sample size is n = 23 making the degrees of freedom df = 22. The significance level is not given sowe take the significance level to be 0.05. Calculating x from the sample data givesx = Σx /n = 241.27 / 23 = 10.49Calculating the sample standard deviation from the data givess = nΣ(x 2 ) − (Σx) 223⋅ 241.27 − (2536.576)2= = 0.507n(n −1)23⋅ 22We move on to the test.H 0 : µ = 10.5H 1 : µ < 10.5The test statistic is t = x − µ 10.49 −10.5=s / n 0.507 / 23 =−0.095In a left-tailed test at the 0.05 significance level with df = 22, the critical value is −t α=−t .05=−1.717.In the row for 22 degrees of freedom, the absolute value for the test statistic falls to the right of 1.321 so the P-value is greater than 0.10.We fail to reject the null hypothesis.There is not sufficient sample evidence to support the claim that the mean winning time in the Olympic 100 mdash is less than 10.5 sec.Notice that the precision of the numbers increases in later years.One important characteristic of the data is that it, generally, decreases in later Olympics. Although the resultsare that there is no evidence that times will be lower than 10.5, it is an invalid conclusion, as they are clearlygoing to continue being lower than 10.5 sec.20. Nicotine in CigarettesThe claim is that the mean nicotine content of cigarettes is greater than 40 mg, so this is a right-tailed test. Thesample size is n = 10 making the degrees of freedom df = 9. The significance level is 0.01. Calculating x fromthe sample data givesx = Σx /n = 433/10 = 43.3Calculating the sample standard deviation from the data givess = nΣ(x 2 ) − (Σx) 210 ⋅18878.9 − (433)2= = 3.801n(n −1)10 ⋅ 9We move on to the test.H 0 : µ = 40H 1 : µ > 40The test statistic is t = x − µs / n=43.3− 40.03.801/ 10 = 2.745In a right-tailed test at the 0.01 significance level with df = 9, the critical value is t α= t .01= 2.821.In the row for 9 degrees of freedom, the test statistic lies between 2.821 and 2.262, so the P-value is between.01 and .025.We fail to reject the null hypothesis.There is not sufficient sample evidence to support the editor’s belief that the mean nicotine content ofcigarettes is greater than 40 mg.21. Norwegian Babies Weigh More?The claim is that Norwegian newborn babies, on average, weigh more than 3420g, which is the mean weightof American newborn babies, so this is a right-tailed test. The sample size is n = 90 making the degrees offreedom df = 89, the sample mean is x = 3570, and the sample standard deviation is s = 498g. Thesignificance level is 0.05.H 0 : µ = 3420H 1 : µ > 3420The test statistic is t = x − µ 3570 − 3420=s / n 498 / 90 = 2.857

Chapter 7: Hypothesis Testing with One Sample 221In a right-tailed test at the 0.05 significance level with df = 89, we use the degrees of freedom closest to 89,which is 90, and so the critical value is t α= t .05= 1.662.In the row for 90 degrees of freedom, the test statistic falls to the left of 2.632 so the P-value is less than 0.005.We reject the null hypothesis.The sample data support the claim that Norwegian newborn babies, on average, weigh more than 3420g, whichis the mean weight of American newborn babies.22. Conductor Life SpanThe claim is that the mean life span of conductors is greater than 69.5 years, which is the mean life span formales in the general population, so this is a right-tailed test. The sample size is n = 35 making the degrees offreedom df = 34, the sample mean is x = 73.4 , and the sample standard deviation is s = 8.7. The significancelevel is 0.05.H 0 : µ = 69.5H 1 : µ > 69.5The test statistic is t = x − µ 73.4 − 69.5=s / n 8.7 / 35 = 2.652In a right-tailed test at the 0.05 significance level with df = 34, the critical value is t α= t .05= 1.691.In the row for 34 degrees of freedom, the test statistic falls between 2.728 and 2.441 so the P-value is between0.005 and 0.01.We reject the null hypothesis.The sample data support the claim that the mean life span of conductors is greater than 69.5 years, which is themean life span for males in the general populationYes, it appears they live longer.Since they become conductors at a later age in life, there are most likely other factors that contribute to theirlongevity.23. Poplar Tree WeightsThe claim is that the fertilized trees come from a population with a mean weight less than 0.92kg, so this is aleft-tailed test. The sample size is n =5 making the degrees of freedom df = 4. The significance level is notmentioned, and so we use 0.05. Calculating x from the sample data givesx = Σx /n = 1.66 /5 = 0.332Calculating the sample standard deviation from the data givess = nΣ(x 2 ) − (Σx) 225⋅1.8284 − (1.66)= = 0.565n(n −1)5⋅ 4We move on to the test.H 0 : µ = 0.92H 1 : µ < 0.92The test statistic is t = x − µ 0.332 − 0.92=s / n 0.565/ 5 =−2.327In a left-tailed test at the 0.05 significance level with df = 4, the critical value is −t α=−t .05=−2.132.In the row for 4 degrees of freedom, the absolute value for the test statistic falls between 2.776 and 2.132 sothe P-value is between 0.025 and 0.05.We reject the null hypothesis.The sample data support the claim that the fertilized trees come from a population with a mean weight lessthan 0.92kg.The fertilizer appears to have an effect, decreasing the weight of poplars on site 1.

222 Chapter 7: Hypothesis Testing with One Sample24. Petal Lengths of IrisesThe claim is that the mean petal length of setosa irises is less than 4.3 mm, the petal length of versicolor irises,so this is a left-tailed test. The sample size is n =50 making the degrees of freedom df = 49. The significancelevel is not mentioned, and so we use 0.05. Calculating x from the sample data givesx = Σx /n = 73.2 /50 = 1.464Calculating the sample standard deviation from the data givess = nΣ(x 2 ) − (Σx) 250 ⋅108.64 − (73.2)2= = 0.1735n(n −1)50 ⋅ 49We move on to the test.H 0 : µ = 4.3H 1 : µ < 4.3The test statistic is t = x − µ 1.464 − 4.3=s / n 0.1735/ 50 =−115.582In a left-tailed test at the 0.05 significance level with df = 49, we use the degrees of freedom closest to 49,which is 50, and so the critical value is −t α=−t .05=−1.676.In the row for 50 degrees of freedom, the absolute value for the test statistic falls far to the left of 2.678 so theP-value is below 0.005 (much lower, one suspects).We reject the null hypothesis.The sample data support the claim that the mean petal length of setosa irises is less than 4.3mm, the petallength of versicolor irises.25. Using the Wrong DistributionIf one uses the standard normal distribution when the Student t distribution is indicated, there is a greaterchance of rejecting the null hypothesis. This is because the critical values found are closer to the center thanwhen dealing with the Student t. This means a test statistic closer to the center could be rejected when, had theStudent t be properly employed, the test statistic would not have fallen in the critical region.26. Finding Critical t ValuesFor a significance level of 0.05, z α= z .05= 1.645. Using this and that the degrees of freedom is df = 9, wecalculateA.z(8 ⋅ df + 3) 1.645(8 ⋅ 9 + 3)A = = = 1.690(8 ⋅ df +1) (8 ⋅ 9 +1)Using this value, we approximate the critical t valuet = df ⋅ (e A 2 / df−1) = 9 ⋅ (e 1.6902 /9−1) = 1.833This value is the same value as displayed in Table A-3.7-6 Testing a Claim About a Standard Deviation of VarianceIn Exercises 1-4, find the test statistic, then use table A-4 to find the critical value(s) of χ 2 and limits containingthe P-value, then determine whether there is sufficient evidence to support the alternate hypothesis.(n1. The test statistic is χ 2 −1)s2=σ 219 ⋅102= = 8.444.15 2The alternate hypothesis is H 1 : σ ≠ 15, so this is a two-tailed test. Since n = 20, df = 19, and the significancelevel is 0.05. The critical values, from Table A-4 are 8.907 and 32.852. Since the test statistic falls between7.733 and 8.907, the P-value is between 0.02 and 0.05. There is sufficient evidence to support the alternatehypothesis.

Chapter 7: Hypothesis Testing with One Sample 2232(n −1)s2. The test statistic is χ 2 =σ 2= 4 ⋅18212 2 = 9.0.The alternate hypothesis is H 1 : σ >15, so this is a right-tailed test. Since n = 5, df = 4, and the significancelevel is 0.01. The critical value, from Table A-4 is 13.277. Since the test statistic falls between 7.779 and9.488, the P-value is between 0.05 and 0.10. There is not sufficient evidence to support the alternatehypothesis.2(n −1)s 29 ⋅3. The test statistic is χ 2 302= = = 10.44 .σ 2 50 2The alternate hypothesis is H 1 : σ

224 Chapter 7: Hypothesis Testing with One Sample6. Supermodel WeightsThe claim is that the population standard deviation of supermodel body weights is less than 29lb, so this is aleft-tailed test. The sample size is n = 9 making the degrees of freedom df = 8. The significance level is 0.01.Calculating the sample standard deviation from the data givess = nΣ(x 2 ) − (Σx) 29 ⋅132,223− (1,089)2= = 7.533n(n −1)9 ⋅8We move on to the test.H 0 : σ = 29H 1 : σ < 29(nThe test statistic is χ 2 −1)s2=σ 2 = 8 ⋅ 7.533229 2 = 0.540In a left-tailed test at the 0.01 significance level with df = 8, the critical value is χ 2 = 2.733.In the row for 8 degrees of freedom, the test statistic lies below 1.344, so the P-value is less than 0.005.We reject the null hypothesis.The sample data support the claim that the population standard deviation of supermodel body weights is lessthan 29lb.7. Supermodel HeightsThe claim is that the population standard deviation of supermodel heights vary less than the heights of womenin general, who have a standard deviation of 2.5 in, so this is a left-tailed test. The sample size is n = 18making the degrees of freedom df = 17. The significance level is 0.05. Calculating the sample standarddeviation from the data givess = nΣ(x 2 ) − (Σx) 218 ⋅88,084 − (1,259)2= = 1.187n(n −1)18 ⋅17We move on to the test.H 0 : σ = 2.5H 1 : σ < 2.52(n −1)s 17The test statistic is χ 2 ⋅1.1872= = = 3.832σ 2 2.5 2In a left-tailed test at the 0.05 significance level with df = 17, the critical value is χ 2 = 8.672.In the row for 17 degrees of freedom, the test statistic lies below 5.697, so the P-value is less than 0.005.We reject the null hypothesis.The sample data support the claim that the population standard deviation of supermodel heights vary less thanthe heights of women in general.8. Using Birth Weight DataThe claim is that the population standard deviation of birth weights of male babies born to mothers taking aspecial vitamin supplement is 0.470 kg, so this is a two-tailed test. The sample size is n = 16 making thedegrees of freedom df = 15. The significance level is not mentioned, so we use 0.05. Calculating the samplestandard deviation from the data givess = nΣ(x 2 ) − (Σx) 2216 ⋅ 222.571− (58.8)= = 0.657n(n −1)16 ⋅15We move on to the test.H 0 : σ = 0.470H 1 : σ 0.470 ≠(nThe test statistic is χ 2 −1)s2 15⋅ 0.6572= = = 29.311σ 2 0.470 2In a two-tailed test at the 0.05 significance level with df = 15, the critical values are χ 2 = 6.262 andχ 2 = 27.488.In the row for 15 degrees of freedom, the test statistic lies between 27.488 and 30.578, so the P-value isbetween 0.05 and 0.02.We reject the null hypothesis.

Chapter 7: Hypothesis Testing with One Sample 225There is sufficient evidence to warrant the rejection of the claim that the population standard deviation of birthweights of male babies born to mothers taking a special vitamin supplement is 0.470 kg.9. Is the New Machine Better?The claim is that the population standard deviation of cold medicine poured by the new machine is less thanthat of the old machine, which was 0.15 oz., so this is a left-tailed test. The sample size is n = 71 making thedegrees of freedom df = 70. The significance level is 0.05. The sample standard deviation is s = 0.12 oz. Wemove on to the test.H 0 : σ = 0.15H 1 : σ < 0.1570 ⋅ 0.1220.15 2 = 44.82(n −1)sThe test statistic is χ 2 = =σ 2In a left-tailed test at the 0.05 significance level with df = 70, the critical value is χ 2 = 51.739.In the row for 70 degrees of freedom, the test statistic lies between 43.275 and 45.442, so the P-value isbetween 0.005 and 0.01.We reject the null hypothesis.The sample data support the claim that the population standard deviation of cold medicine poured by the newmachine is less than that of the old machine. Its purchase should be considered.10. Systolic Blood Pressure for WomenWe use the pre-exercise, no stress systolic blood pressure for this hypothesis test. The claim is that thepopulation standard deviation of systolic blood pressure for women is 23.4, so this is a two-tailed test. Thesample size is n = 40 making the degrees of freedom df = 39. The significance level is not mentioned, so weuse 0.05. Calculating the sample standard deviation from the data givess = nΣ(x 2 ) − (Σx) 240 ⋅ 502,488 − (4432)2= = 17.114n(n −1)40 ⋅ 39We move on to the test.H 0 : σ = 23.4H 1 : σ 23.4 ≠22(n −1)s 39 ⋅17.114The test statistic is χ 2 = = = 20.861σ 2 23.4 2The degrees of freedom df = 39 is not displayed in Table A-4, so we use df = 40. In a two-tailed test at the 0.05significance level with df = 40, the critical values are χ 2 = 24.433 and χ 2 = 59.342.In the row for 40 degrees of freedom, the test statistic lies between 20.707 and 22.164, so the P-value isbetween 0.01 and 0.02.We reject the null hypothesis.There is sufficient evidence to warrant rejection of the claim that the population standard deviation of systolicblood pressure for women is 23.4.

226 Chapter 7: Hypothesis Testing with One Sample11. Weights of MenThe claim is that the population standard deviation of weights for men is 28.7 lb., so this is a two-tailed test.The sample size is n = 40 making the degrees of freedom df = 39. The significance level is 0.05. Calculatingthe sample standard deviation from the data givess = nΣ(x 2 ) − (Σx) 240 ⋅1,217,971.76 − (6,902.0)2= = 26.327n(n −1)40 ⋅ 39We move on to the test.H 0 : σ = 28.7H 1 ≠ : σ 28.72(n −1)s 39 ⋅The test statistic is χ 2 26.3272= = = 32.817σ 2 28.7 2The degrees of freedom df = 39 is not displayed in Table A-4,so we use df = 40. In a two-tailed test at the 0.05significance level with df = 40, the critical values are χ 2 = 24.433 and χ 2 = 59.342.In the row for 40 degrees of freedom, the test statistic lies between 29.051 and 51.805, so the P-value is largerthan 0.20.We fail to reject the null hypothesis.There is not sufficient evidence to warrant rejection of the claim that the population standard deviation ofweights for men is 28.7 lb. If the standard deviation was larger, the maximum load displayed in an elevatormay be inaccurate.12. Heights of WomenThe claim is that the population standard deviation of heights for women is 2.52 in., so this is a two-tailed test.The sample size is n = 40 making the degrees of freedom df = 39. The significance level is 0.05. Calculatingthe sample standard deviation from the data givess = nΣ(x 2 ) − (Σx) 240 ⋅160,037.38 − (2,527.8)2= = 2.741n(n −1)40 ⋅ 39We move on to the test.H 0 : σ = 2.52H 1 ≠ : σ 2.522(n −1)s 39 ⋅The test statistic is χ 2 2.7412= = = 46.140σ 2 2.52 2The degrees of freedom df = 39 is not displayed in Table A-4,so we use df = 40. In a two-tailed test at the 0.05significance level with df = 40, the critical values are χ 2 = 24.433 and χ 2 = 59.342.In the row for 40 degrees of freedom, the test statistic lies between 29.051 and 51.805, so the P-value is largerthan 0.20.We fail to reject the null hypothesis.There is not sufficient evidence to warrant rejection of the claim that the population standard deviation ofheights for women is 2.52 in. If the standard deviation was larger, the typical heights range for women wouldbe larger, and a greater number of women would have safety belts that were not fitted properly.13. Finding Critical Values for χ 2a. For α = 0.05, the critical values found on Table A-2 are z = -1.96 and z = 1.96. Also, k = degrees offreedom = n – 1 = 100. Calculating the left critical value for a χ 2 , we find( ) 2 = 73.772.χ 2 = 1 2 (z + 2k −1) 2 = 1 −1.96 + 2 ×100 −12Calculating the right critical value for χ 2 , we findχ 2 = 1 2 (z + 2k −1) 2 = 1 ( 1.96 + 2 ×100 −1) 2 = 129.0702The values from Table A-4 are 74.222 and 129.561. The values found using the formula are fairly accurate(each is off by roughly 0.5).

Chapter 7: Hypothesis Testing with One Sample 227b. For α = 0.05, the critical values found on Table A-2 are z = -1.96 and z = 1.96. Also, k = degrees offreedom = n – 1 = 149. Calculating the left critical value for a χ 2 , we find( ) 2 = 116.643.χ 2 = 1 2 (z + 2k −1) 2 = 1 −1.96 + 2 ×149 −12Calculating the right critical value for χ 2 , we findχ 2 = 1 2 (z + 2k −1) 2 = 1 1.96 + 2 ×149 −12( ) 2 = 184.19914. Finding Critical Values for χ 2a. For α = 0.05, the critical values found on Table A-2 are z = -1.96 and z = 1.96. Also, k = degrees offreedom = n – 1 = 100. Calculating the left critical value for a χ 2 , we findχ 2 = k 1− 29k + z 29k3= 100 1−29 ×100 −1.96 29 ×100Calculating the right critical value for χ 2 , we findχ 2 = k 1− 29k + z 29k3= 100 1−29 ×100 +1.96 29 ×10033= 74.216.= 129.565The values from Table A-4 are 74.222 and 129.561. The values found using the formula are extremelyaccurate (each is off by less than 0.01).b. For α = 0.05, the critical values found on Table A-2 are z = -1.96 and z = 1.96. Also, k = degrees offreedom = n – 1 = 149. Calculating the left critical value for a χ 2 , we findχ 2 = k 1− 29k + z 29k3= 149 1−29 ×149 −1.96 29 ×149Calculating the right critical value for χ 2 , we findχ 2 = k 1− 29k + z 29k3= 149 1−29 ×149 +1.96 29 ×14933= 117.093.= 184.690Review Exercises1. a. We can conclude that most adult Americans are opposed to estate taxes, but we need to be careful. Withsuch a large sample size, a sample proportion that is only slightly higher than 0.50 would result in thatconclusion.b. This drug should not be used. The actual mean weight loss is insignificant, though it does have statisticalsignificance.c. The 0.001 significance level would be most convincing. However, I would prefer the 0.01 significancelevel, as it is a reasonable threshold without being unrealistic.d. There is not sufficient evidence to warrant the rejection of the claim that the volume of cold medicine isdifferent than 12 oz.e. “rejecting a true null hypothesis.”2. Identifying Hypotheses and Distributionsa. H 1 : µ < 10,000. Since the sample size is larger than 30 (n = 750) and the population standard deviation isunknown, the sampling distribution is the Student t distribution.b. H 1 : σ > 1.8. We do not know the distribution of the population, and so we cannot use the χ 2 distribution.However, had we known the distribution was normally distributed, we would use the distribution, sa weare performing a hypothesis test regarding the population standard deviation.c. H 1 : p > 0.50. Since np = 500 × 0.50 = 250 and nq = 500 × 0.50 = 250, the sampling distribution is thenormal distribution.d. H 1 : µ ≠ 100. Since the sample size is larger than 30 (n = 150) and the population standard deviation isknown, the sampling distribution is the normal distribution.

228 Chapter 7: Hypothesis Testing with One Sample3. Effective TreatmentFirst, we check the requirements. We will assume that the sample is a simple random sample. The conditionsfor a binomial are satisfied. The sample size is 200 with the claim that the proportion of tree deaths for treatedtrees is less than 10%, making p = 0.10, so np = 200 × 0.10 = 20 and nq = 200 × 0.90 = 180. The requirementsare satisfied.The claim is that the proportion of tree deaths for treated trees is less than 10%, so this is a left-tailed test. Inthe sample, 16 of the trees died the following year. The sample proportion is p ˆ = x / n = 16 / 200 = 0.08.H 0 : p = 0.10H 1 : p < 0.10The test statistic is z = p ˆ − p 0.08 − 0.10= =−0.943pq 0.10 × 0.90n 200In a left-tailed test at the 0.05 significance level, the critical value is −z α=−z .05=−1.645.Since this is a left tailed test, the P-value is the area to the left of the test statistic. Using Table A-2, we findthat the P − value = 0.1736.We fail to reject the null hypothesis.There is not sufficient sample evidence to support the claim that the proportion of tree deaths for treated treesis less than 10%.4. X-Linked Recessive DisorderWe will test the claim that the method has lowered the proportion of boys who have the X-linked recessivedisorder. First, we check the requirements. We will assume that the sample is a simple random sample. Theconditions for a binomial are satisfied. The sample size is 40 with the claim that the proportion of boys whohave the X-linked recessive disorder is less than 50%, making p = 0.50, so np = 40 × 0.50 = 20 andnq = 40 × 0.50 = 20. The requirements are satisfied.The claim is that the proportion of boys who have the X-linked recessive disorder is less than 50%, so this is aleft-tailed test. In the sample, 15 of the boys had the disorder. The sample proportion isp ˆ = x / n = 15/ 40 = 0.375.H 0 : p = 0.50H 1 : p < 0.50The test statistic is z = p ˆ − ppqn0.375− 0.50= =−1.5810.50 × 0.5040In a left-tailed test at the 0.05 significance level, the critical value is −z α=−z .05=−1.645.Since this is a left tailed test, the P-value is the area to the left of the test statistic. Using Table A-2, we findthat the P − value = 0.0571.We fail to reject the null hypothesis.There is not sufficient sample evidence to support the claim that the proportion of boys who have the X-linkedrecessive disorder is less than 50%. Though the sample proportion was below the 50%, there is not sufficientevidence to make this claim.5. Are Consumers Being Cheated?The claim is that the mean volume of alcohol in the bottles is less than 12 oz, so this is a left-tailed test. Thesample size is n = 24, the sample mean is x = 11.4 , and the sample standard deviation is s = 0.62. Thesignificance level is not specified, so we use 0.05.H 0 : µ = 12H 1 : µ < 12The test statistic is t = x − µ 11.4 −12.0=s / n 0.62 / 24 =−4.741

Chapter 7: Hypothesis Testing with One Sample 229In a left-tailed test at the 0.01 significance level with 23 degrees of freedom, the critical value ist α= t .05= 1.714 .In the row for 23 degrees of freedom, the absolute value for the test statistic is greater than 2.807 so the P-value is less than 0.005.We reject the null hypothesis.The sample data support the claim that the mean volume of alcohol in the bottles is less than 12 oz. However,the claim by Harry Windsor has a little merit. However, the sample size is close to 30, so if the distribution issymmetric, the hypothesis test would still be valid.6. Controlling CholesterolThe sample is a random sample. We will assume that the sample is from a normally distributed population.The claim is that the mean amount of the drug is 25 mg, so this is a two-tailed test. The sample size is n = 15making the degrees of freedom df = 14. The significance level is 0.05. Calculating x from the sample datagivesx = Σx /n = 373.3/15 = 24.887Calculating the sample standard deviation from the data givess = nΣ(x 2 ) − (Σx) 215⋅ 9,295.07 − (373.3)2= = 0.590n(n −1)15⋅14We move on to the test.H 0 : µ = 25H 1 : µ ≠ 25The test statistic is t = x − µ 24.887 − 25=s / n 0.590 / 15 =−0.742In a two-tailed test at the 0.05 significance level with df = 14, the critical values are ±t α=±t .025=±2.145.In the row for 14 degrees of freedom, the absolute value of the test statistic is less than 1.345, so the P-value isgreater than 0.20.We fail to reject the null hypothesis.There is not sufficient evidence to warrant the rejection of the claim that the mean amount of the drug is 25mg.7. Controlling VariationThe sample is a random sample. We will assume that the sample is from a normally distributed population.The claim is that the population standard deviation for the drug is greater than 0.5 mg, so this is a two-tailedtest. The sample size is n = 15 making the degrees of freedom df = 14. The significance level is 0.05.Calculating the sample standard deviation from the data givess = nΣ(x 2 ) − (Σx) 215⋅ 9,295.07 − (373.3)2= = 0.590n(n −1)15⋅14We move on to the test.H 0 : σ = 0.5H 1 : σ > 0.52(n −1)s 14 ⋅The test statistic is χ 2 0.5902= = = 19.494σ 2 0.5 2In a right-tailed test at the 0.05 significance level with df = 14, the critical value is χ 2 = 23.685.In the row for 14 degrees of freedom, the test statistic lies between 7.790 and 21.064, so the P-value is greaterthan 0.10 but less than 0.90.We fail to reject the null hypothesis.There is not sufficient evidence to warrant the rejection of the claim that the population standard deviation forthe drug is greater than 0.5 mg. If the variance is too great, then the amount of drug in some of the pills may betoo large, may exceed safe levels.

230 Chapter 7: Hypothesis Testing with One Sample8. Effectiveness of New Chicken FeedIt is reasonable to assume that the data comes form a normally distributed population. The claim is that themean weight of newborn chickens using the new feed is greater than 62.2 oz, so this is a right-tailed test. Thesample size is n = 9 making the degrees of freedom df = 8. The significance level is 0.01. Calculating x fromthe sample data givesx = Σx /n = 579.4 /9 = 64.378Calculating the sample standard deviation from the data givess = nΣ(x 2 ) − (Σx) 29 ⋅ 37,334.6 − (579.4)2= = 2.065n(n −1)9 ⋅8We move on to the test.H 0 : µ = 62.2H 1 : µ > 62.2The test statistic is t = x − µ 64.378 − 62.2= = 3.164s / n 2.065/ 9In a right-tailed test at the 0.01 significance level with df = 8, the critical value is t α= t .01= 2.896.In the row for 8 degrees of freedom, the test statistic lies between 3.250 and 2.821, so the P-value is between0.005 and 0.01.We reject the null hypothesis.The sample data support the clam that the mean weight of newborn chickens using the new feed is greater than62.2 oz.9. Effectiveness of Seat BeltsThe sample size is larger than 30, so using the Student t distribution is appropriate. The claim is that the meanamount of time spent in an intensive care unit for children, who were wearing safety belts, following anautomobile accident is less than 1.39 days, so this is a left-tailed test. The sample size is n = 123 making thedegrees of freedom df = 122, the sample mean is x = 0.83, and the sample standard deviation is s = 0.16. Thesignificance level is 0.01.H 0 : µ = 1.39H 1 : µ < 1.39The test statistic is t = x − µs / n = 0.83−1.390.16 / 123 =−38.817The critical values for 122 degrees of freedom is not on the table, so we use df = 100. In a left-tailed test at the0.01 significance level with df = 100, the critical value is −t α=−t .01=−2.364.In the row for 100 degrees of freedom, the absolute value of the test statistic falls far to the left of 2.626 so theP-value is less than 0.005.We reject the null hypothesis.The sample data support the claim that the mean amount of time spent in an intensive care unit for children,who were wearing safety belts, following an automobile accident is less than 1.39 days, which is the meannumber of days spent following an accident when children are not using a safety belt. There is strong evidenceshowing that seatbelts seem to help.

Chapter 7: Hypothesis Testing with One Sample 231Cumulative Review Exercises1. Monitoring DioxinHere is the data listed in order, and the necessary sums.0.018 0.0268 0.0281 0.032 0.044 0.0524 0.161 0.175 0.176Σx = 0.7133 Σx 2 = 0.09506a. The mean is x = Σx /n = 0.7133/ 9 = 0.07926b. The median is 0.044c. The standard deviation is s = nΣ(x 2 ) − (Σx) 29 ⋅ 0.09506− (0.7133)2= = 0.0694n(n −1)9 ⋅8d. The variance is s 2 = nΣ(x 2 ) − (Σx) 29 ⋅ 0.09506 − (0.7133)2= = 0.00482n(n −1)9 ⋅8e. The range is (0.176-0.018) = 0.158f. We assume that the data is from a normally distributed population. For a 95% confidence interval, α=.05,and so α/2=.025. Since n=9, the degrees of freedom df=n–1=8. Using Table A-3, we see that t .025= 2.306.The margin of error is E = t .025The confidence interval isx − E < µ < x + Esn= 2.3060.06949= 0.05330.07926 − 0.05335 < µ < 0.07926 + 0.053350.02591 < µ < 0.13261g. We again assume that the sample data come from a normally distributed population. The claim is that themean dioxin level is less than 0.16 ng/m 3 , so this is a left-tailed test. The sample size is n = 9 making thedegrees of freedom df = 8, the sample mean is x = 0.07926, and the sample standard deviation is s =0.0694. The significance level is 0.05.H 0 : µ = 0.16H 1 : µ < 0.16The test statistic is t = x − µ 0.07926− 0.16= =−3.490s / n 0.0694 / 9In a left-tailed test at the 0.05 significance level with df = 8, the critical value is −t α=−t .05=−1.860.In the row for 8 degrees of freedom, the absolute value for the test statistic is greater than 3.355 so the P-value is less than 0.005.We reject the null hypothesis.The sample data support the claim that the mean dioxin level is less than 0.16 ng/m 3 .h. One important characteristic of the data that is not addressed is that there was a dramatic drop in valuesover time, in that the 4 th and later measures were all much lower than the initial three measurements.

232 Chapter 7: Hypothesis Testing with One Sample2. Weights of Babiesa. This is the area to the right of 3000 under the normal curve with mean µ = 3420 and standard deviation σ =495. In order to use table A-2, we must convert into the standard normal distribution,z = x − µ 3000 − 3420= =−0.85.σ 495The probability is the area to the right of -0.85, and so the probability is P = 1 – 0.1977 = 0.8023.b. Since the babies are independent of one another, the probability that all five have weights greater than 3000g is P = (0.8023) 5 = 0.3324 .c. We use the central limit theorem to convert to the standard normal distribution. The probability will be thearea to the right of 3000 under the normal curve with mean µ x= µ = 3420 and standard deviationσ x= σ / n = 495/ 5 = 221.371.z = x − µ 3000 − 3420x= =−1.90.σ x221.371The probability is the area to the right of -1.90, and so the probability is P = 1 – 0.0287 = 0.9713.d. Looking for the area 0.90 in the body of Table A-2, we find the z score to be 1.28. We findP 90= µ + z ⋅σ = 3420 +1.28 ⋅ 495 = 4053.6.3. Blood Pressure Readingsa. The mean is x = Σx /n = 1987.7 /16 = 124.231The standard deviation is s = nΣ(x 2 ) − (Σx) 2216 ⋅ 254543.43− (1987.7)= = 22.523n(n −1)16 ⋅15b. The sample is a random sample. We will assume that the sample is from a normally distributed population.The claim is that the mean systolic blood pressure for women infected with the new viral strain is 114.8, sothis is a two-tailed test. The sample size is n = 16 making the degrees of freedom df = 15. The significancelevel is 0.05.H 0 : µ = 114.8H 1 : µ 114.8 ≠The test statistic is t = x − µs / n = 124.231−114.8 = 1.67522.523/ 16In a two-tailed test at the 0.05 significance level with df = 15, the critical values are ±t α=±t .025=±2.131.In the row for 15 degrees of freedom, the test statistic is between 1.341 and 1.753, so the P-value isbetween 0.10 and 0.20.We fail to reject the null hypothesis.There is not sufficient evidence to warrant the rejection of the claim that the mean systolic blood pressurefor women infected with the new viral strain is 114.8.c. We again assume that the data is from a normally distributed population. For a 95% confidence interval,α=.05, and so α/2=.025. Since n=16, the degrees of freedom df=n–1=15. Using Table A-3, we see thatt .025= 2.131.The margin of error is E = t .025The confidence interval isx − E < µ < x + Esn222.523= 2.131 = 11.99916124.231−11.999 < µ < 124.231+11.999112.232 < µ < 136.23The confidence interval limits do contain the value 114.8

Chapter 7: Hypothesis Testing with One Sample 233d. The claim is that the population standard deviation of systolic blood pressure of women infected with thenew viral strain is 13.1, so this is a two-tailed test. The sample size is n = 16 making the degrees of freedomdf = 15. The significance level is 0.05.H 0 : σ = 13.1H 1 : ≠ σ 13.12(n −1)s 15⋅The test statistic is χ 2 22.5232= = = 44.341σ 2 13.1 2In a two-tailed test at the 0.05 significance level with df = 15, the critical values are χ 2 = 6.262 andχ 2 = 27.488.In the row for 15 degrees of freedom, the test statistic lies above 32.801, so the P-value is less than 0.01.We reject the null hypothesis.There is sufficient evidence to warrant the rejection of the claim that the population standard deviation ofsystolic blood pressure of women infected with the new viral strain is 13.1.e. Yes, the variability of the blood pressures has greatly increased.