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. $h_{0}:mult21.62$ choose the correct alternative hypothesis: 1. $h_{a}:mu
eq21.62$ 2. $h_{a}:mu = 21.62$ 3. $h_{a}:mult21.62$ 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) +/ - 2.02 (to 3 decimals), the null hypothesis.

Explanation:

Step1: Calculate the sample mean and standard deviation

Assume the data in the file is used to calculate the sample mean $\bar{x}$ and sample standard deviation $s$.

Step2: Calculate the test statistic $t$

The formula for the test statistic in a one - sample $t$ - test is $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$, where $\mu = 21.62$ (the population mean under the null hypothesis), $n = 42$ (sample size).

Step3: Determine the degrees of freedom

The degrees of freedom for a one - sample $t$ - test is $df=n - 1=42-1 = 41$.

Step4: Find the $t$ value

Using the formula $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$ with the calculated $\bar{x}$ and $s$ from the data (not provided here, but assuming standard calculation steps), we get the $t$ value.

Answer:

Assuming the data in the file is used to calculate the sample mean $\bar{x}$ and sample standard deviation $s$, and using the formula $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$ with $\mu = 21.62$ and $n = 42$, the $t$ value (to 3 decimals) is calculated. For example, if $\bar{x}=22$, $s = 3$:

$$t=\frac{22 - 21.62}{3/\sqrt{42}}\approx\frac{0.38}{3/6.48}\approx\frac{0.38}{0.463}\approx0.821$$

So the $t$ value (to 3 decimals) is $0.821$ (this is a sample calculation, actual value depends on data in the file).