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A study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois. Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids. Steroids, which are dangerous, are sometimes used in an attempt to improve athletic performance. Researchers want to know if there's a difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids. Perform an appropriate hypothesis test using the 4-Step Process.

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Final answer:

To perform an appropriate hypothesis test using the 4-Step Process, we need to determine if there is a difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids.

Step-by-step explanation:

To perform an appropriate hypothesis test using the 4-Step Process, we need to determine if there is a difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids. The null hypothesis, denoted as H0, assumes there is no difference in the proportions. The alternative hypothesis, denoted as Ha, assumes there is a difference in the proportions. The 4-Step Process involves: 1. Stating the hypotheses, 2. Formulating an analysis plan, 3. Analyzing sample data, and 4. Interpreting the results. Based on the given information, we can use a two-sample z-test for proportions to analyze the data and determine if there is a significant difference in the proportion of freshmen and seniors who have used anabolic steroids.

User Geethanga
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6 votes

Answer:

We conclude that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids.

Step-by-step explanation:

We are given that a study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois.

Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids.

Let
p_1 = proportion of Illinois high school freshmen who have used anabolic steroids.


p_2 = proportion of Illinois high school seniors who have used anabolic steroids.

SO, Null Hypothesis,
H_0 :
p_1=p_2 {means that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids}

Alternate Hypothesis,
H_A :
p_1\\eq p_2 {means that there is a significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids}

The test statistics that would be used here Two-sample z test for proportions;

T.S. =
\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } } ~ N(0,1)

where,
\hat p_1 = sample proportion of high school freshmen who have used anabolic steroids =
(34)/(1679) = 0.0203


\hat p_2 = sample proportion of high school seniors who have used anabolic steroids =
(24)/(1366) = 0.0176


n_1 = sample of high school freshmen = 1679


n_2 = sample of high school seniors = 1366

So, the test statistics =
\frac{(0.0203-0.0176)-(0)}{\sqrt{(0.0203(1-0.0203))/(1679)+(0.0176(1-0.0176))/(1366) } }

= 0.545

The value of z test statistics is 0.545.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids.

User Plannapus
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