Question

You may need to use the appropriate appendix table or technology to answer this question. DATAfile: ResidentialWater On its municipal website, the city of Tulsa states that the rate it charges per 5 CCF of residential water is $21.62. How do the residential water rates of other U.S. public utilities compare to Tulsa's rate? The data shown below ($) contains the rate per 5 CCF of residential water for 42 randomly selected U.S. cities. 10.48 9.18 11.8 6.5 12.42 14.53 15.56 10.12 14.5 16.18 17.6 19.18 17.98 12.85 16.8 17.35 15.64 14.8 18.91 17.99 14.9 18.42 16.05 26.85 22.32 22.76 20.98 23.45 19.05 23.7 19.26 23.75 27.8 27.05 27.14 26.99 24.68 37.86 26.51 39.01 29.46 41.65 (a) Formulate hypotheses that can be used to determine whether the population mean rate per 5 CCF of residential water charged by U.S. public utilities differs from the $21.62 rate charged by Tulsa. (Enter != for * as needed.) HO: u= 21.62 Ha u! = 21.62 (b) What is the test statistic for your hypothesis test in part (a)? (Round your answer to three decimal places.) -1.147 What is the p-value for your hypothesis test in part (a)? (Round your answer to four decimal places.) (c) At a = 0.05, can your null hypothesis be rejected? What is your conclusion? O Do not reject Ho. The mean rate per 5 CCF of residential water throughout the United States differs significantly from the rate per 5 CCF of residential water in Tulsa. O Reject H. The mean rate per 5 CCF of residential water throughout the United States does not differ significantly from the rate per 5 CCF of residential water in Tulsa. O Reject H. The mean rate per 5 CCF of residential water throughout the United States differs significantly from the rate per 5 CCF of residential water in Tulsa. Do not reject Ho. The mean rate per 5 CCF of residential water throughout the United States does not differ significantly from the rate per 5 CCF of residential water in Tulsa.

Answer 1

The provided sample mean is X = 20.23833 and the sample standard deviation is s = 7.7991, and the sample size is n 42. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u = 21.62 Ha: u 7 21.62 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 = 2.02. The rejection region for this two-tailed test is R = {t:t > 2.02} (3) Test Statistics The t-statistic is computed as follows: 20.23833 – 21.62 t= X μο s/n = -1.148 || 7.7991/V 42

t-statistic = -1.14811 =-1.148


(4) Decision about the null hypothesis Since it is observed that t = 1.148 <te = 2.02, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p= 0.2576, and since p = 0.2576 > 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean u is different than 21.62, at the 0.05 significance level.

p-value = 0.2576
That are the only two answers you needed.