Chapter 7. Hypothesis Testing with One Sample

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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
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Chapter 7. Hypothesis Testing with One Sample

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