Chapter 10.1, Problem 17E

a.

To determine

To conduct: The significance test at 5% level of significance.

Answer to Problem 17E

There is sufficient evidence to conclude that the difference exists between population proportions.

Explanation of Solution

Given:

  $x_1=632,n_1=800,x_2=268,n_2=400$

Formula used:

The test statistic formula is:

  $z=\frac{\left({\widehat{p}}_1-{\widehat{p}}_2\right)-\left(p_1-p_2\right)}{\sqrt{\widehat{p}\left(1-\widehat{p}\right)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}$

The formula to compute the confidence interval for difference in population proportions is:

  $CI=\left({\widehat{p}}_1-{\widehat{p}}_2\right)\pm z_{\alpha /2}\times \sqrt{\frac{{\widehat{p}}_1\left(1-{\widehat{p}}_1\right)}{n_1}+\frac{{\widehat{p}}_2\left(1-{\widehat{p}}_2\right)}{n_2}}$

Calculation:

The null and alternative hypotheses are:

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1\ne p_2\end{array}$

The output using Ti-83 calculator is:

  

Decision rule:

The p-value is less than the significance level. Thus, decision is written as the rejection of the null hypothesis.

Conclusion:

At 5% significance level, there are sufficient evidence to support the provided claim of difference between the population proportions.

b.

To determine

To find: The 95% confidence interval for the differences in the proportion. Also, comment on the consistency of the above part.

Answer to Problem 17E

The confidence interval is (0.066, 0.174)

Explanation of Solution

Calculation:

The 95% confidence interval using Ti-83 plus calculator is computed as:

  

Thus, the confidence interval is (0.066, 0.174)

Interpretation:

Thus, it could be interpreted that there is 95% probability that the difference in the proportion of lies between 0.066 and 0.174.

Here, 0 is not lying in the computed confidence interval. Thus, it could be said that the confidence interval is consistent with the result of above part.