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 file ResidentialWater contains the rate per 5 CCF of residential water for 42 randomly selected U.S. cities.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.b.What is the p-value for your hypothesis test in part (a)?c.At a=.05, can your null hypothesis be rejected? What is your conclusion?d.Repeat the preceding hypothesis test using the critical value approach.

c) At a significance level of 0.05, the P-value is greater. In this case, the effect is not significant. The null hypothesis is failed to be rejected.

The conclusion is that there is no enough statistical evidence to tell that the cities of the U.S has a different rate than Tulsa.

d) For this significance level, the critical values for t are t=±2.019.

In this situation, with a statistic t=-1.15, it falls in the acceptance region, so the null hypothesis is not rejected.

Step-by-step explanation:

The question is incomplete:

The list of rates per 5 CCF for 42 randomly selected U.S. cities has a sample mean of 20.24 and a standard deviation of 7.80.

a) The null hypothesis should state that the population mean rate is equal to the one from Tulsa. The alternative hypothesis should state they differ.

[tex]H_0: \mu=21.62\\\\H_a: \mu\neq21.62[/tex]

b) The degrees of freedom are:

[tex]df=n-1=42-1=41[/tex]

The t-statistic can be calculated as:

[tex]t=\frac{M-\mu}{s/\sqrt{n}} =\frac{20.24-21.62}{7.80/\sqrt{42}} =\frac{-1.38}{1.20}=-1.15[/tex]

The P-value of t is:

[tex]p-value=2P(t<-1.15)=2*0.1284=0.2568[/tex]