**Answer:**

a)

b) P-value=0.2568

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:**

<em>The question is incomplete:</em>

<em>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.</em>

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.

b) The degrees of freedom are:

The t-statistic can be calculated as:

The P-value of t is: