Chapter 10.1, Problem 20E

(a)

To determine

To explain why the sample results give evidence for the alternative hypothesis.

Explanation of Solution

This problem refers to the previous question, that is,

  $\begin{array}{l}{x}_1=34\\ x_2=24\\ n_1=1679\\ n_2=1366\end{array}$

It is given in the question that the researchers want to know that if there is a difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids or not.

So, the given claim that: difference between the proportions.

Now, we have to find out the appropriate hypotheses for performing a significance test.

Thus, the claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population proportions are equal. If the null hypothesis is the claim then the alternative hypothesis states the opposite of the null hypothesis.

Therefore, the appropriate hypotheses for this is:

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

Where we have,

  $p_1=$ the proportion of high school freshmen in Illinois that used anabolic steroids.

  $p_2=$ the proportion of high school seniors in Illinois that used anabolic steroids.

Now, the sample proportion is the number of successes divided by the sample size, that is:

  $\begin{array}{l}{\widehat{p}}_1=\frac{x_1}{n_1}\\ =\frac{34}{1679}\\ =0.02025\\ {\widehat{p}}_2=\frac{x_2}{n_2}\\ =\frac{24}{1366}\\ =0.01757\end{array}$

From this we conclude that the sample proportion for the second sample is different than the sample proportion for the first sample which is same as the alternative hypothesis i.e. $p_1\ne p_2$ and thus this implies that there is some evidence for the alternative hypothesis. This states that the proportion of high school freshmen in Illinois that used anabolic steroids is different from the proportion of high school seniors in Illinois that used anabolic steroids.

(b)

To determine

To calculate the standardized test statistics and the P -value.

Answer to Problem 20E

The P-value is $0.6600$ and the value of test statistics value is $0.44$ .

Explanation of Solution

Now referring to the part (a), we know that:

  $\begin{array}{l}{x}_1=34\\ x_2=24\\ n_1=1679\\ n_2=1366\end{array}$

And the appropriate hypotheses for this is:

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

Where we have,

  $p_1=$ the proportion of high school freshmen in Illinois that used anabolic steroids.

  $p_2=$ the proportion of high school seniors in Illinois that used anabolic steroids.

And the sample proportion is the number of successes divided by the sample size, that is:

  $\begin{array}{l}{\widehat{p}}_1=\frac{x_1}{n_1}\\ =\frac{34}{1679}\\ =0.02025\\ {\widehat{p}}_2=\frac{x_2}{n_2}\\ =\frac{24}{1366}\\ =0.01757\end{array}$

  $\begin{array}{l}{\widehat{p}}_p=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{34+24}{1679+1366}\\ =\frac{58}{2045}\\ =0.02836\end{array}$

Now, we will calculate the value of test statistics as:

  $\begin{array}{l}z=\frac{{\widehat{p}}_1-{\widehat{p}}_2-(p_1-p_2)}{\sqrt{{\widehat{p}}_p(1-{\widehat{p}}_p)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\\ =\frac{0.02025-0.01757-0}{\sqrt{0.02836(1-0.02836)}\sqrt{\frac{1}{1679}+\frac{1}{1366}}}\\ \approx 0.44\end{array}$

The P-value is the probability of obtaining the value of the test statistics or a value more extreme assuming that the null hypothesis is true. Thus, we have,

  $\begin{array}{l}P=P(Z<-0.44\text{ or }Z>0.44)\\ =2P(Z<-0.44)\\ =2(0.3300)\\ =0.6600\end{array}$

Therefore, the P-value is $0.6600$ and the value of test statistics value is $0.44$ .

(c)

To determine

To explain what conclusion would you make.

Answer to Problem 20E

There is no convincing evidence that the proportion of high school freshmen in Illinois that used anabolic steroids is different from the proportion of high school seniors in Illinois that used anabolic steroids.

Explanation of Solution

It is given that:

  $\begin{array}{l}{x}_1=34\\ x_2=24\\ n_1=1679\\ n_2=1366\end{array}$

And we calculated in part (a) and part (b), that the P-value is $0.6600$ and the value of test statistics value is $0.44$ .

And the hypotheses are as:

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

Thus, if the P-value is smaller than the significance level, then we will reject the null hypothesis, thus, we have,

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

Thus, we conclude that there is no convincing evidence that the proportion of high school freshmen in Illinois that used anabolic steroids is different from the proportion of high school seniors in Illinois that used anabolic steroids.