Chapter 26, Problem 40E

a)

To determine

To find the z-value

Answer to Problem 40E

Z-statistic is 21.2953

Explanation of Solution

Given:

  

Formula:

Sample proportion:

  $\widehat{p}=\frac{x}{n}$

Z-statistic:

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

First, need to calculate sample proportions.

  ${\widehat{p}}_1=\frac{343}{710}=0.4831$

  ${\widehat{p}}_2=\frac{127}{1491}=0.0852$

Pooled proportion is,

  $\overline{p}=\frac{343+127}{710+1491}=0.2135$

Therefore, z-statistic is,

  $z=\frac{0.4831-0.0852}{\sqrt{0.2135(1-0.2135)\left(\frac{1}{710}+\frac{1}{1491}\right)}}=21.2953$

Therefore, z=21.2953

Now,

  $z^2={21.2953}^2=453.49\approx {\chi }^2$

b)

To determine

To show that square of z value is equal to calculated chi-square.

Answer to Problem 40E

The square of z values is approximately equal to chi-square value.

Explanation of Solution

Given:

  

z=21.2953

  ${\chi }^2$ = 453.48 and P-value = 0

Therefore, z=21.2953

Now,

  $z^2={21.2953}^2=453.49\approx {\chi }^2$

Hence, square of z values is approximately equal to chi-square value.

c)

To determine

To show that p-value from z statistic and chi-square test are same.

Answer to Problem 40E

Both p-values are almost same.

Explanation of Solution

Given:

z=21.2953

  ${\chi }^2$ = 453.48 and P-value = 0

Z-statistic is 21.2953

Using, excel formula,

=2*(1-NORMSDIST(21.2953))

P-value = 0

Both p-values are almost same.