380 CHAPTER 7 Hypothesis Testing with One Sample Hypothesis Testing Using a Rejection Region A used car dealer says that the mean listing price of all used vehicles sold in the past 12 months is at least $23,500. You suspect this claim is incorrect and find that a random sample of 14 used vehicles sold in the past 12 months has a mean listing price of $21,558 and a standard deviation of $3350. Is there enough evidence to reject the dealer’s claim at a = 0.05? Assume the population is normally distributed. (Adapted from Edmunds.com) SOLUTION Because s is unknown, the sample is random, and the population is normally distributed, you can use the t@test. The claim is “the mean listing price is at least $23,500.” So, the null and alternative hypotheses are H0: m Ú $23,500 (Claim) and Ha: m 6 $23,500. The test is a left-tailed test, the level of significance is a = 0.05, and the degrees of freedom are d.f. = 14 - 1 = 13. So, using Table 5, the critical value is t0 = -1.771. The rejection region is t 6 -1.771. The standardized test statistic is t = x - m s 1n Because s is unknown and the population is normally distributed, use the t@test. = 21,558 - 23,500 3350 214 Assume m = 23,500. ≈ -2.169. Round to three decimal places. The figure shows the location of the rejection region and the standardized test statistic t. Because t is in the rejection region, you reject the null hypothesis. Interpretation There is enough evidence at the 5% level of significance to reject the claim that the mean listing price of all used vehicles sold in the past 12 months is at least $23,500. TRY IT YOURSELF 4 An industry analyst says that the mean transaction price of all new vehicles sold in the past 12 months is less than $43,500. A random sample of 25 new vehicles sold in the past 12 months has a mean transaction price of $40,573 and a standard deviation of $6250. Is there enough evidence to support the analyst’s claim at a = 0.10? Assume the population is normally distributed. (Adapted from Edmunds.com) Answer: Page A41 Remember that when you make a decision, the possibility of a type I or a type II error exists. For instance, in Example 4, a type I error is possible when you reject H0, because m Ú $23,500 may be true. 5% Level of Significance t −3 −2 −1 0 1 2 3 t ≈ −2.169 t0 = −1.771 α= 0.05 To explore this topic further, see Activity 7.3 on page 386. 7.3 See Minitab steps on page 414. EXAMPLE 4
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