Question

Steroids in high school Refer to Exercise 16.

a. Explain why the sample results give some evidence for the alternative hypothesis.

b. Calculate the standardized test statistic and P-value.

c. What conclusion would you make?

Short answer

a. $p_1\ne p_2$and evidence for alternative hypothesis is present.

b. The $P$value is $0.6600$and test statistics value is $0.44$.

c. There is no evidence that 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.

Step-by-step solution

Given Information

It is given that $x_1=34$

$x_2=24$

$n_1=1679$

$n_2=1366$

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

Why Sample results give evidence for Alternative Hypothesis

Claim is difference between proportions.

The appropriate hypothesis:

Null: $H_0:p_1=p_2$

Alternative: $H_a:p_1\ne p_2$

$p_1$is proportion of high school freshmen in Illinois that used anabolic steroids and $p_2$is proportion of high school seniors in Illinois that used anabolic steroids.

Sample proportion is role="math" localid="1654710840637" ${\hat{p}}_1=\frac{x_1}{n_1}=\frac{34}{1679}=0.02025$

and ${\hat{p}}_2=\frac{x_2}{n_2}=\frac{24}{1366}=0.01757$

$p_1\ne p_2$ and 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.

P value and test statistics

As ${\hat{p}}_1=0.02025$and ${\hat{p}}_2=0.01757$

and ${\hat{p}}_p=\frac{x_1+x_2}{n_1+n_2}=\frac{34+24}{1679+1366}=0.02836$

Value of test statistics is

$z=\frac{{\hat{p}}_1-{\hat{p}}_2-\left(p_1-p_2\right)}{\sqrt{{\hat{p}}_p\left(1-{\hat{p}}_p\right)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

role="math" localid="1654711122778" $=\frac{0.02025-0.01757-0}{\sqrt{0.02836(1-0.02836)}\sqrt{\frac{1}{1679}+\frac{1}{1366}}}\approx 0.44$

Probability is $P=P(Z<-0.44\text{or}Z>0.44)$

$=2P(Z<-0.44)$

$=2(0.3300)$

$=0.6600$

Conclusion

From given data and above calculations

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

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.