Question

A physical education teacher wanted to test whether her county’s students are falling below the national proportion. It is known that the national proportion for high school juniors passing the aerobic portion of the test (running a mile in less than 10 minutes) is 0.70. Using α = 0.05, is there sufficient evidence to conclude that less than 70% of students run one mile in less than 10 minutes? Conduct a full hypothesis test by following the steps below.

a. Define the population parameter in context in one sentence.
b. State the null and alternative hypotheses using correct notation.
c. State the significance level for this problem.
d. Check the three conditions of the Central Limit Theorem that allow you to use the one- proportion z-test using one complete sentence for each condition. Show work for the numerical calculation. You can assume the population is large.
e. Calculate the test statistic "by-hand." Show the work necessary to obtain the value by typing all the steps needed and the resulting test statistic. Do not round while doing the calculation. Then, round the test statistic to decimal places.
f. Calculate the p-value using the standard Normal table and provide the answer. Use four decimal places for the p-value.
g. State whether you reject or do not reject the null hypothesis and the reason for your decision in one sentence.
h. Based on your above decision, state your conclusion in context of the problem (i.e. interpret your results and/or answer the question being posed) in one or two complete sentences.

Answer 1

(a) The population parameter is students who ran the mile in less than 10 minutes.

(b) The hypothesis being tested is:

H0: p = 0.70

Ha: p < 0.70

(c) The significance level is 0.05.

(d) In order to conduct a one-sample proportion z-test, the following conditions should be met:

The data are a simple random sample from the population of interest.

The population is at least 10 times as large as the sample.

n⋅p ≥ 10 and n⋅(1−p) ≥ 10, where n is the sample size and p is the true population proportion.

n⋅p = 756 * 0.70 ≥ 10

n⋅(1−p) = 756 * (1 - 0.70) ≥ 10

All the conditions are met.

(e) p = 504/756 = 0.67

The test statistic, z = (p - p)/√p(1-p)/n = (0.67 - 0.70)/√0.70(1-0.70)/756 = -2.00

(f) The p-value is 0.0228.

(g) Since the p-value (0.0228) is less than the significance level (0.05), we can reject the null hypothesis.

(h) Therefore, we can conclude that less than 70% of students run one mile in less than 10 minutes.

Explanation: