Chapter 7. Hypothesis Testing with One Sample

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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.
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Chapter 7. Hypothesis Testing with One Sample

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