Question

8. Listed below are the heights (inches) for the simple random sample of supermodels. Use a 0.05 significance level to test the claim that supermodels have heights with a mean that is greater than the mean height of 63.8 in. for women in the general population. Do not use the p-value. 70 71 69.25 68.5 69 70 71 70 70 69.5

Answer 1

The provided sample mean is X = 69.83 and the sample standard deviation is s = 0.8, and the sample size is n = 10. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u = 63.8 Ha: u> 63.8 This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a right-tailed testis te = 1.833. The rejection region for this right-tailed testis R=t:t> 1.833 (3) Test Statistics The t-statistic is computed as follows: X - μo t= 69.83 - 63.8 0.8/10 = 23.836 s/n (4) Decision about the null hypothesis Since it is observed that t = 23.836 > te = 1.833, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p=0, and since p=0 < 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean u is greater than 63.8, at the 0.05 significance level. Confidence Interval The 95% confidence interval is 69.258 < u < 70.402.

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