Question

A report states the average rent for office space in a urban area is $17.75 per square foot. A real estate agent claims this average is incorrect. The agent selected a sample of 28 rental properties and found their mean to be $19.25 per square foot, with a sample standard deviation of $3.55 per square foot. Test the claim at alpha = 0.05. Use the P-value method to evaluate/compare with the given alpha (0.05).

For question #18, give your calculated value for "t",rounding to the hundredths place.

For question #19 (multiple choice), choose between what two areas your calculated t-value (when converted to P-value) falls.

For question #20, answer True or False with regard to rejecting the null hypothesis.  Hint: this is a two-tailed test!!

Answer 1

The provided sample mean is X = 19.25 and the sample standard deviation is 3.55, and the sample size is n = 28. S = (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u = 17.75 Ha: u 7 17.75 This corresponds to a two-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 two-tailed test is te = 2.052. The rejection region for this two-tailed test is R= {t: 1t| > 2.052} (3) Test Statistics The t-statistic is computed as follows: t= - HO s/n 19.25 – 17.75 3.55/28 - 2.236 (4) Decision about the null hypothesis Since it is observed that t1 = 2.236 > tc = 2.052, it is then concluded that the null hypothesis is rejected. = 0.0338 Using the P-value approach: The p-value is p = 0.0338, and since p 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 different than 17.75, at the 0.05 significance level. Confidence Interval The 95% confidence interval is 17.873 < p < 20.627. Graphically. T-Test Results: t-stats = 2.236, p-value =0.0338 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 -4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

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