Question

on its municipal website, the city of tulsa states that the rate it charges per 6 ccf of residential water is $21.62. how do the residential water rates of other u.s. public utilities compare to tulsas rate? the file residentialwater contains the rate per 6 ccf of residential water for 42 randomly selected u.s. cities. click on the datafile logo to reference the data. data file a. formulate hypotheses that can be used to determine whether the population mean rate per 6 ccf of residential water charged by u.s. public utilities differs from the $21.62 rate charged by tulsa. choose the correct null hypothesis: 1. $h_{0}:mu = 21.62$ 2. $h_{0}:mu
eq21.62$ 3. $h_{0}:mu < 21.62$ choose the correct alternative hypothesis: 1. $h_{a}:mu
eq21.62$ 2. $h_{a}:mu = 21.62$ 3. $h_{a}:mu < 21.62$ b. what is the p - value for your hypothesis test in part (a)? round your answer to four decimal places. c. at $alpha = 0.05$, can your null hypothesis be rejected? what is your conclusion? the null hypothesis. the mean rate per 6 ccf of residential water throughout the u.s. significantly from the rate per 6 ccf of residential water in tulsa. d. repeat the preceding hypothesis test using the critical value approach. the critical value(s) is(are) (to 3 decimals), the null hypothesis.

Explanation:

Step1: Hypothesis Formulation

The null hypothesis \(H_0\) assumes no difference. So \(H_0:\mu = 21.62\). The alternative hypothesis \(H_a\) for a difference (two - tailed test) is \(H_a:\mu
eq21.62\).

Step2: P - value Interpretation

A P - value of \(0.2575>0.05\) (significance level \(\alpha = 0.05\)). In hypothesis testing, if \(P-\text{value}>\alpha\), we do not reject the null hypothesis.

Step3: Critical Value Approach

For a two - tailed test with \(\alpha = 0.05\) and \(n-1=42 - 1 = 41\) degrees of freedom (approximate using \(t\) - distribution), the critical values are \(\pm t_{\alpha/2,n - 1}\approx\pm2.02\). The test statistic \(t = 0.822\). Since \(|t|=0.822<2.02\), we do not reject the null hypothesis.

Answer:

a. Null hypothesis: \(H_0:\mu = 21.62\); Alternative hypothesis: \(H_a:\mu
eq21.62\)
b. \(0.2575\)
c. Do not reject the null hypothesis. The mean rate per \(6\) CCF of residential water throughout the U.S. does not differ significantly from the rate per \(6\) CCF of residential water in Tulsa.
d. Critical values: \(\pm2.02\); Test statistic \(t = 0.822\); Do not reject the null hypothesis.