Chapter 20, Problem 20.21E

(a)

To determine

To explain why we cannot safely use either the large-sample confidence interval or the test for comparing the proportions of normal and altered mice that develop tumors.

Explanation of Solution

In the question, it is given that out of the $\text{33}$ mice with lowered levels of DNA methylation, $\text{23}$ developed tumors. None of the control group of $\text{18}$ normal mice developed tumors in the same time period. Thus, we cannot safely use either the large-sample confidence interval or the test for comparing the proportions of normal and altered mice that develop tumors because the sample proportion for control group is less than five and we cannot create large sample confidence interval. Only the large sample confidence interval cannot be safely used since the sample is not large enough and the test for comparing proportions can be used.

(b)

To determine

To find out what are the sample sizes and the numbers of mice with tumors after you do this and give a plus four $\text{99\%}$ confidence interval for the difference in the proportions of the two populations that develop tumors.

Answer to Problem 20.21E

The sample sizes are $\text{35 and 20}$ and the numbers of mice with tumors are $0.6857\text{ and 0}\text{.05}$ and the plus four $\text{99\%}$ confidence interval is $(0.398,0.874)$ .

Explanation of Solution

In the question, it is given that out of the $\text{33}$ mice with lowered levels of DNA methylation, $\text{23}$ developed tumors. None of the control group of $\text{18}$ normal mice developed tumors in the same time period. The plus four method adds two observations, a success and a failure to each sample. Thus, the sample sizes and the numbers of mice with tumors after we do this are as:

  $\begin{array}{l}{\tilde{p}}_1=\frac{x_1+1}{n_1+2}\\ =\frac{23+1}{33+2}\\ =0.6857\\ {\tilde{p}}_2=\frac{x_2+1}{n_2+2}\\ =\frac{0+1}{18+2}\\ =0.05\end{array}$

The sample sizes are:

  $\begin{array}{l}{n}_1+2=33+2\\ =35\\ n_2+2=18+2\\ =20\end{array}$

Thus, plus four $\text{99\%}$ confidence interval for the difference in the proportions of the two populations that develop tumors is calculated as:

  $\begin{array}{l}SE=\sqrt{\frac{{\tilde{p}}_1(1-{\tilde{p}}_1)}{n_1+2}+\frac{{\tilde{p}}_2(1-{\tilde{p}}_2)}{n_2+2}}\\ =\sqrt{\frac{0.6857(1-0.6857)}{33+2}+\frac{0.05(1-0.05)}{18+2}}\\ =0.09237\\ z_{\alpha /2}=z_{0.01/2}=2.576\\ (0.6857-0.05)\pm 2.576\times 0.09237\\ =0.6357\pm 0.238\\ =(0.398,0.874)\end{array}$

(c)

To determine

To explain based on your confidence interval that is the difference between normal and altered mice significant at the $\text{1\%}$ level.

Answer to Problem 20.21E

Yes, the difference is significant at the $\text{1\%}$ level.

Explanation of Solution

In the question, it is given that out of the $\text{33}$ mice with lowered levels of DNA methylation, $\text{23}$ developed tumors. None of the control group of $\text{18}$ normal mice developed tumors in the same time period. The plus four method adds two observations, a success and a failure to each sample. Thus, the sample sizes and the numbers of mice with tumors after we do this are as:

  $\begin{array}{l}{\tilde{p}}_1=\frac{x_1+1}{n_1+2}\\ =\frac{23+1}{33+2}\\ =0.6857\\ {\tilde{p}}_2=\frac{x_2+1}{n_2+2}\\ =\frac{0+1}{18+2}\\ =0.05\end{array}$

Thus, plus four $\text{99\%}$ confidence interval for the difference in the proportions of the two populations that develop tumors is calculated as:

  $\begin{array}{l}SE=\sqrt{\frac{{\tilde{p}}_1(1-{\tilde{p}}_1)}{n_1+2}+\frac{{\tilde{p}}_2(1-{\tilde{p}}_2)}{n_2+2}}\\ =\sqrt{\frac{0.6857(1-0.6857)}{33+2}+\frac{0.05(1-0.05)}{18+2}}\\ =0.09237\\ z_{\alpha /2}=z_{0.01/2}=2.576\\ (0.6857-0.05)\pm 2.576\times 0.09237\\ =0.6357\pm 0.238\\ =(0.398,0.874)\end{array}$

Thus, we can see that in the plus four $\text{99\%}$ confidence interval for the difference in the proportions of the two populations that develop tumors does not contain zero thus the difference is significant at $\text{1\%}$ significance level.