Chapter 9.2, Problem 9.32E

a)

To determine

To perform hypothesis test for one sample proportion $\widehat{p}$ .

Explanation of Solution

Given:

Null and alternative hypothesis:

  $\begin{array}{l}{H}_o:p=0.68\\ H_a:p>0.68\end{array}$

Test statistic = z = 2.35

Calculation:

Test statistic z = 2.35

Using the P-value approach:

P(Z>2.35) = 1-P(Z<2.35)

P(Z>2.35) = 1-0.9906-=0.0094

Using excel formula, =NORMSDIST(2.35)

The p-value is p = 0.0094

If a = 0.05 then p = 0.0094<0.05, the result is significant.

If a = 0.01 then p = 0.0094<0.01, the result is significant.

b)

To determine

To perform hypothesis test for one sample proportion $\widehat{p}$ .

Explanation of Solution

Given:

Null and alternative hypothesis:

  $\begin{array}{l}{H}_o:p=0.68\\ H_a:p\ne 0.68\end{array}$

Test statistic = z = 2.35

Calculation:

Test statistic z = 2.35

Using the P-value approach:

  $2\times P\left(Z<2.35\right)=2\times 0.0094=0.0188$

Using excel formula, =NORMSDIST(2.35)

The p-value is p = 0.0188

If a = 0.05 then p = 0.0188 <0.05, the result is significant.

If a = 0.01 then p = 0.0188 >0.01, the result is insignificant.