Question

In 2012, researchers at Pennsylvania State University discovered adult Americans exercise an average of 17 minutes per day. To test if the amount of exercise in New York City is more than the national average, a researcher decides to do a hypothesis test, at a 1% significance level. He surveys 50 New Yorkers randomly and asks them about their amount of exercise each day, on average. From the data, the sample mean time (x¯) is 20 minutes per day, and the sample standard deviation (s) is 1.6 minutes. H0: μ=17; Ha: μ>17 α = 0.01 (significance level) What is the test statistic (t-value) of this one-mean hypothesis test (with σ unknown)?

Answer 1

t ≈ 13.26

Explanation:

We are given;

x¯ = 20 minutes per day

μ = 17 minutes per day

sample standard deviation; s = 1.6 minutes

Sample size; n = 50

When σ is unknown, formula for test statistic is;

t = (x¯ - μ)/(s/√n)

t = (20 - 17)/(1.6/√50)

t = 3/0.2263

t ≈ 13.26