SECTION 7.3 Hypothesis Testing for the Mean (s Unknown) 381 Hypothesis Testing Using Rejection Regions An industrial company claims that the mean pH level of the water in a nearby river is 6.8. You randomly select 39 water samples and measure the pH of each. The sample mean and standard deviation are 6.7 and 0.35, respectively. Is there enough evidence to reject the company’s claim at a = 0.05? SOLUTION Because s is unknown, the sample is random, and n = 39 Ú 30, you can use the t@test. The claim is “the mean pH level is 6.8.” So, the null and alternative hypotheses are H0: m = 6.8 (Claim) and Ha: m ≠ 6.8. The test is a two-tailed test, the level of significance is a = 0.05, and the degrees of freedom are d.f. = 39 - 1 = 38. So, using Table 5, the critical values are -t0 = -2.024 and t0 = 2.024. The rejection regions are t 6 -2.024 and t 7 2.024. The standardized test statistic is t = x - m s 1n Because s is unknown and n Ú 30, use the t@test. = 6.7 - 6.8 0.35 239 Assume m = 6.8. ≈ -1.784. Round to three decimal places. The figure shows the location of the rejection regions and the standardized test statistic t. Because t is not in the rejection region, you fail to reject the null hypothesis. You can confirm this decision using technology, as shown below. MINITAB One-Sample T N Mean StDev SE Mean 95% CI for μ T-Value P-Value 39 6.7000 0.3500 0.0560 (6.5865, 6.8135) -1.78 0.082 μ: population mean of Sample Null hypothesis H0: μ = 6.8 Alternative hypothesis H1: μ ≠ 6.8 Interpretation There is not enough evidence at the 5% level of significance to reject the claim that the mean pH level is 6.8. TRY IT YOURSELF 5 The company in Example 5 claims that the mean conductivity of the river is 1890 milligrams per liter. The conductivity of a water sample is a measure of the total dissolved solids in the sample. You randomly select 39 water samples and measure the conductivity of each. The sample mean and standard deviation are 2350 milligrams per liter and 900 milligrams per liter, respectively. Is there enough evidence to reject the company’s claim at a = 0.01? Answer: Page A41 See TI-84 Plus steps on page 415. EXAMPLE 5 t0 = 2.024 −t0 = −2.024 t ≈ −1.784 t −3 −4 −1 0 1 2 3 4 α= 0.025 α= 0.025 1 2 1 2 5% Level of Significance
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