Chapter 22, Problem 26E

(a)

To determine

To Explain: that these outcomes help the effectiveness of routine mammograms in reducing deaths from breast cancer.

Answer to Problem 26E

  $\begin{array}{c}{H}_0:p_1-p_2=0\\ H_A:p_1-p_2>0\end{array}$

Explanation of Solution

Given:

  $\begin{array}{c}{y}_1=66\\ n_1=21195\\ y_2=63\\ n_2=21088\end{array}$

Formula used:

  $\begin{array}{c}{\widehat{p}}_{Pooled}=\frac{y_1+y_2}{n_1+n_2}\\ S{E}_{Pooled}\left({\widehat{p}}_1-{\widehat{p}}_2\right)=\sqrt{\frac{{\widehat{p}}_{Pooled}{\widehat{q}}_{Pooled}}{n_1}+\frac{{\widehat{p}}_{Pooled}{\widehat{q}}_{Pooled}}{n_2}}\end{array}$

Calculation:

  $H_0:$ In order to prevent breast cancer deaths, mammograms might not be an appropriate screening tool.

  $i.e,H_0:p_1-p_2=0$

  $H_A:$ Mammograms can be an important method for screening to minimise deaths from breast cancer.

  $i.e,H_A:p_1-p_2>0$

Here, $p_1$ shows the percentage of deaths that have never had mammograms and ; $p_2$ shows the percentage of deaths that have been tested for mammograms.

  $\begin{array}{c}{\widehat{p}}_1=\frac{y_1}{n_1}\\ =\frac{66}{21195}\\ =0.0031\\ {\widehat{p}}_2=\frac{y_2}{n_2}\\ =\frac{63}{21088}\\ =0.0030\end{array}$

The pooled proportion is,

  $\begin{array}{c}{\widehat{p}}_{Pooled}=\frac{y_1+y_2}{n_1+n_2}\\ =\frac{66+63}{21195+21088}\\ =0.0031\end{array}$

The pooled value to estimate the standard error is,

  $\begin{array}{c}S{E}_{Pooled}\left({\widehat{p}}_1-{\widehat{p}}_2\right)=\sqrt{\frac{{\widehat{p}}_{Pooled}{\widehat{q}}_{Pooled}}{n_1}+\frac{{\widehat{p}}_{Pooled}{\widehat{q}}_{Pooled}}{n_2}}\\ =\sqrt{\frac{\left(0.0031\right)\left(0.9969\right)}{21195}+\frac{\left(0.0031\right)\left(0.9969\right)}{21088}}\\ =0.0005\end{array}$

The observed difference in proportions is

  $\begin{array}{c}{\widehat{p}}_1-{\widehat{p}}_2=0.0031-0.0030\\ =0.0001\end{array}$

The test statistic is,

  $\begin{array}{c}z=\frac{\left({\widehat{p}}_1-{\widehat{p}}_2\right)-0}{S{E}_{Pooled}\left({\widehat{p}}_1-{\widehat{p}}_2\right)}\\ =\frac{0.0001}{0.0005}\\ =0.20\end{array}$

This is a right tailed test, so the p-value is,

  $\begin{array}{c}P=P\left(z>0.20\right)\\ =1-P\left(z\le 0.20\right)\\ =1-0.5793\\ =0.4207\end{array}$

Since the P-value is very high, null hypotheses are not dismissed. There is insufficient evidence to suggest that mammograms could be an important method for screening to minimise deaths from breast cancer.

(b)

To determine

To Explain: if conclusion is false, types of error have that committed.

Explanation of Solution

It will make an error with type I. It states that there is a false null hypothesis, but there is refuse to reject it.