Chapter PVI, Problem 16RE

(a)

To determine

To create a $95\%$ confidence interval for the change in enrolment rates.

Answer to Problem 16RE

We are $95\%$ confident that the change in enrolment rates lie between $(-0.0819,0.0419)$ .

Explanation of Solution

It is given in the question that in September, CNN reported on a method of grad student recruiting by the Haas school of Business at U.C.-Berkeley. Thus, to find the confidence interval for proportion we will use the calculator $\text{TI}-89$ . To calculate a confidence interval for a population proportion: Go to the STAT Intsmenu and select $6:2-$ PropZInt. Enter the observed counts and the sample size for both the samples. Specify a confidence level. Calculate the interval.So, we have,

  $\begin{array}{l}{p}_1=0.52\\ n_1=500\\ p_2=0.54\\ n_2=500\end{array}$

And $\alpha =0.05$ so by using the calculator $\text{TI}-89$ , the confidence interval is as:

  $(-0.0819,0.0419)$ .

So, we are $95\%$ confident that the change in enrolment rates lie between $(-0.0819,0.0419)$ .

(b)

To determine

To explain are you convinced that this new form of recruiting has been effective.

Answer to Problem 16RE

Yes, we are convinced this new form of recruiting has been effective.

Explanation of Solution

It is given in the question that in September, CNN reported on a method of grad student recruiting by the Haas school of Business at U.C.-Berkeley.And $\alpha =0.05$ so by using the calculator $\text{TI}-89$ , the confidence interval is as:

  $(-0.0819,0.0419)$ .

So, we are $95\%$ confident that the change in enrolment rates lie between $(-0.0819,0.0419)$ . Thus, since zero lie in this interval then we are convinced that this new form of recruiting has been effective.