Answer:
(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.
Step-by-step explanation:
would be your mixed fraction.
