Chapter 10.1, Problem 18E

Solution

a)

To verify the claim that there is a significance difference between the proportions of freshmen and seniors who used anabolic steroids.

The sample data does not give good evidence that there is a significance difference between the proportions of freshmen and seniors who used anabolic steroids.

Given:sample size of high school freshmen = 1679

sample size of high school seniors = 1366

Freshmen who used steroids = 34

Seniors who used steroids = 24

Concept used:The conditions to satisfy to run one significance test

1) The sample should be random

2) np = 10

3) n(1-p) = 10

4) There should be at most 10% of people of population in the sample.

Formula used:$\begin{array}{l}{\hat{p}}_c=\frac{x_1+x_2}{n_1+n_2}\\ Z=\frac{\left({\hat{p}}_1-{\hat{p}}_2\right)-0}{\sqrt{\frac{{\hat{p}}_c(1-{\hat{p}}_c)}{n_1}+\frac{{\hat{p}}_c(1-{\hat{p}}_c)}{n_2}}}\end{array}$

Calculation:

Let P₁ be proportion of freshmen who used anabolic steroids.

P₂ be proportion of seniors who used anabolic steroids.

The null and alternate hypothesis are shown below

  $\begin{array}{l}{H}_0:p_1-p_2=0\\ H_a:p_1-p_2\ne 0\end{array}$

The sample of freshmen and seniors are drawn randomly.

The sample size is 1679 freshmen and 1366 seniors, there will be at least 16790 freshmen and 13660 seniors in total. So the 10% condition is also met.

The proportion of freshmen and seniors in sample who use steroids are calculated as shown below

  $\begin{array}{l}\widehat{p_1}=\frac{34}{1679}=0.0203\\ \widehat{p_2}=\frac{24}{1366}=0.0176\end{array}$

  $\begin{array}{l}{n}_1\widehat{p_1}=1679*0.0203=34\\ n_1\widehat{p_1}>10\\ n_1(1-\widehat{p_1})=1679*(1-0.0203)=1645\\ n_1(1-\widehat{p_1})>10\\ n_2{\hat{p}}_2=1366*0.0176=24\\ n_2\widehat{p_2}>10\\ n_2(1-\widehat{p_2})=1366*(1-0.0176)=1342\\ n_2(1-{\hat{p}}_2)>10\end{array}$

All conditions are met so we can use significance test.

The pooled proportion is calculated as shown below

  ${\hat{p}}_c=\frac{x_1+x_2}{n_1+n_2}=\frac{34+24}{1679+1366}=0.0190$

The test statistic is calculated as shown below

  $\begin{array}{l}Z=\frac{\left({\hat{p}}_1-{\hat{p}}_2\right)-0}{\sqrt{\frac{{\hat{p}}_c(1-{\hat{p}}_c)}{n_1}+\frac{{\hat{p}}_c(1-{\hat{p}}_c)}{n_2}}}\\ Z=\frac{\left(0.0203-0.0176\right)-0}{\sqrt{\frac{0.019*0.981}{1679}+\frac{0.019*0.981}{1366}}}=0.54\end{array}$

This is two tail test and the p value is 2*P(Z = 0.54), with respect to table we get p value is 2*0.2946 which is 0.5892.

The p value is greater than the level of significance 0.05. So at 5% level of significance there is no evidence to reject the null hypothesis. So, there is no evidence to conclude that there is significant difference in actual proportions of freshmen and seniors in who use anabolic steroids.

b)

To calculate the 95% confidence interval for difference between the proportions of freshmen and seniors who used anabolic steroids.

The 95% confidence interval for difference between the proportions of freshmen and seniors who used anabolic steroids is (-0.007, 0.0124)

Given:The proportion of Freshmen who use anabolic steroids $p_1=0.0203$

The proportion of Seniors who use anabolic steroids $p_2=0.0176$

The number of Freshmen who use anabolic steroids $n_1=\text{ 1679}$

The number of Seniors who use anabolic steroids $n_2=\text{ 1}366$

Formula used:$ConfidenceInterval=\left({\hat{p}}_1-{\hat{p}}_2\right)\pm Z*\sqrt{\frac{{\hat{p}}_1(1-{\hat{p}}_1)}{n_1}+\frac{{\hat{p}}_2(1-{\hat{p}}_2)}{n_2}}$

Calculation:

Substituting the values we get the following

The Z critical value for 95% confidence interval is 1.96

  $\begin{array}{l}ConfidenceInterval=\left({\hat{p}}_1-{\hat{p}}_2\right)\pm Z*\sqrt{\frac{{\hat{p}}_1(1-{\hat{p}}_1)}{n_1}+\frac{{\hat{p}}_2(1-{\hat{p}}_2)}{n_2}}\\ LowerInterval=\left({\hat{p}}_1-{\hat{p}}_2\right)-Z*\sqrt{\frac{{\hat{p}}_1(1-{\hat{p}}_1)}{n_1}+\frac{{\hat{p}}_2(1-{\hat{p}}_2)}{n_2}}\\ LowerInterval=\left(0.0203-0.0176\right)-1.96*\sqrt{\frac{0.0203*0.9797}{1679}+\frac{0.0176*0.9824}{1366}}\\ LowerInterval=0.0027-0.0097=-0.007\\ UpperInterval=\left({\hat{p}}_1-{\hat{p}}_2\right)+Z*\sqrt{\frac{{\hat{p}}_1(1-{\hat{p}}_1)}{n_1}+\frac{{\hat{p}}_2(1-{\hat{p}}_2)}{n_2}}\\ LowerInterval=\left(0.0203-0.0176\right)+1.96*\sqrt{\frac{0.0203*0.9797}{1679}+\frac{0.0176*0.9824}{1366}}\\ LowerInterval=0.0027+0.0097=0.0124\end{array}$

Conclusion:We are 95% confident that the interval from -0.007 to 0.0124 has the true difference in proportions of freshmen and seniors who use anabolic steroids. As 0 is possible value for the difference of proportions we can conclude that there is no significant difference between the proportions of freshmen and seniors who use anabolic steroids.