SECTION 7.5 Hypothesis Testing for Variance and Standard Deviation 399 Using a Hypothesis Test for the Population Variance A sporting goods manufacturer claims that the variance of the strengths of a certain fishing line is 15.9. A random sample of 15 fishing line spools has a variance of 21.8. At a = 0.05, is there enough evidence to reject the manufacturer’s claim? Assume the population is normally distributed. SOLUTION Because the sample is random and the population is normally distributed, you can use the chi-square test. The claim is “the variance is 15.9.” So, the null and alternative hypotheses are H0: s 2 = 15.9 (Claim) and Ha: s 2 ≠ 15.9. The test is a two-tailed test, the level of significance is a = 0.05, and the degrees of freedom are d.f. = 15 - 1 = 14. Using Table 6, the critical values are x 2 L = 5.629 and x 2 R = 26.119. The rejection regions are x 2 6 5.629 and x 2 7 26.119. The standardized test statistic is x 2 = 1n - 12s2 s 2 Use the chi-square test. = 115 - 12121.82 115.92 Assume s 2 = 15.9. ≈ 19.195. Round to three decimal places. The figure below shows the location of the rejection regions and the standardized test statistic x 2. Because x 2 is not in the rejection regions, you fail to reject the null hypothesis. 20 25 1 2 30 5 10 15 R 2χ α = 26.119 2χ ≈ 19.195 = 0.025 L 2χ = 5.629 1 2 α= 0.025 2χ Interpretation There is not enough evidence at the 5% level of significance to reject the claim that the variance of the strengths of the fishing line is 15.9. TRY IT YOURSELF 6 A company that offers dieting products and weight loss services claims that the variance of the weight losses of their users is 25.5. A random sample of 13 users has a variance of 10.8. At a = 0.10, is there enough evidence to reject the company’s claim? Assume the population is normally distributed. Answer: Page A41 Picturing the World A community center claims that the chlorine level in its pool has a standard deviation of 0.46 parts per million (ppm). A sampling of the pool’s chlorine levels at 25 random times during a month yields a standard deviation of 0.61 ppm. (Adapted from American Pool Supply) Frequency Chlorine level (ppm) 4 3 2 1 1.0 1.4 1.8 2.2 2.6 3.0 x f At 0.05, is there enough evidence to reject the claim? EXAMPLE 6
RkJQdWJsaXNoZXIy NjM5ODQ=