Question

A recent article in Healthy Life magazine claimed that the mean amount of leisure time per week for European

men is 48.2 hours. The distribution of leisure time amounts is reported to be approximately Normal.

You believe the figure of 48.2 is too large and decide to conduct your own test. In a random sample of 25

European men, you find that the mean is 44.6 hours of leisure per week and that the standard deviation of the

sample is 6.9 hours.

(a) State the null hypothesis and the alternative hypothesis for your test.

(b) Conduct the test and determine the p-value.

At 0.01 significance level, can you conclude that the information in the article is untrue?

(c) How would your answer to part (b) be affected if the distribution of leisure time amounts was uniform?

Answer 1

(a) Let denotes the mean leisure time per week for European

men.

We have to test, H0: = 48.2 against H1: < 48.2, where,

H0 is the null hypothesis and H1 is the alternative hypothesis.

(b) The test statistic is given by,

T = , where, = 44.6, = 48.2, s = 6.9, n = 25

Thus, T = - 2.6087. Under H0, T ~ t-distribution with (n-1) i.e.

24 d.f.

The p-value = P(T < - 2.6087) = 0.0077.

Since, p-value < level of significance = 0.01, we reject H0 and

conclude that the information in the article is untrue.

(c) Had the distribution been uniform, we could not apply one

sample t-test. If the sample size is large enough to assume

Normal approximation, then only we can apply the above test.