Question

A large clinical trial of the effect of diet on breast cancer assigned women at random to either a normal diet or a low-fat diet. To check that the random assignment did produce comparable groups, we can compare the two groups at the start of the study. Ask if there is a family history of breast cancer: $3396$of the $19,541$women in the low-fat group and $4929$of the $29,294$women in the control group said “Yes.” If the random assignment worked well, there should not be a significant difference in the proportions with a family history of breast cancer.

(a) How significant is the observed difference? Carry out an appropriate test to help answer this question.

(b) Describe a Type I and a Type II error in this setting. Which is more serious? Explain.

Short answer

(a) There is not sufficient evidence to support the claim of a difference.

(b) Type II error is worse, because it would be assumed that the people were randomly assigned to each group and each group seems to be the same, while it is not the case.

Step-by-step solution

Part(a) Step 1: Given Information

Given

$x_1=3396$

$n_1=19541$

$x_2=4929$

role="math" $n_2=29294$

Determine the hypothesis

$H_0:p_1-p_2=0$

$H_a:p_1-p_2\ne 0$

Part(a) Step 2: Explanation

The sample proportion is the number of successes divided by the sample size:

${\hat{p}}_1=\frac{x_1}{n_1}=\frac{3396}{19541}\approx 0.174$

${\hat{p}}_2=\frac{x_2}{n_2}=\frac{4929}{29294}\approx 0.168$

${\hat{p}}_p=\frac{x_1+x_2}{n_1+n_2}=\frac{3396+4929}{19541+29294}\approx 0.170$

Determine the value of the test statistic:

localid="1650450817963" $$\begin{gathered} z=\frac{{\hat{p}}_1-{\hat{p}}_2}{\sqrt{{\hat{p}}_p\left(1-{\hat{p}}_p\right)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \\ =\frac{0.174-0.168}{\sqrt{0.168(1-0.168)}\sqrt{\frac{1}{19541}+\frac{1}{29294}}} \\ \approx 1.74 \end{gathered}$$

The $p$-value is the probability of obtaining the value of the test statistic, or a value more extreme. Determine the $p$-value using table A:

localid="1650450836161" $$\begin{gathered} P=P(Z<-1.74\text{or}Z>1.74) \\ =2\times P(Z<-1.74) \\ =2\times 0.0409 \\ =0.0818 \end{gathered}$$

If the $p$-value is smaller than the significance level, reject the null hypothesis:

$P>0.05\Rightarrow \text{Fail to reject}{H}_0$

Part(b) Step 1: given Information

Given

$x_1=3396$

$n_1=19541$

$x_2=4929$

$n_2=29294$

Determine the hypothesis

$H_0:p_1-p_2=0$

$H_a:p_1-p_2\ne 0$

Part(b) Step 2: Explanation

(b) Type I error: Reject the null hypothesis $H_0$, when $H_0$is true.

As a result, there is no significant difference between the two groups, even if it appears that there is. Type II error: Failing to reject the null hypothesis$H_0$, when $H_0$is false.

As a result, there is a considerable difference in both groups, even if it looks that there is none. As a result, while the design appears random, it is not.

The Type Il mistake is even more serious.