Question

A study is conducted to look at the time students exercise in average. In a random sample of 115 students, a researcher finds that the mean time students exercise is 11.3hrs/month with a standard deviation of 6.43hrs/month. Using the Critical Value Methodand a 5% significance level, can a researcher conclude that in average students exercise less than 15 hours per month?

Answer 1

The researcher wants to test whether the average exercise time in students is less than 15 hours or not. The hypotheses are :

H0 : the average exercise time in students is equal to 15 hours , i.e , = 15

H1 : the average exercise time in students is less than 15 hours , i.e , < 15

The sample mean , = 11.3 and sample standard deviation , s = 6.43

n = 115

level of significance , alpha = 0.05

The test statistic , t = ( - ) / (s / n)

= ( 11.3 - 15 ) / ( 0.599)

= -6.177

Degrees of freedom = n - 1 = 114

We find the critical value of t using the t table for df = 114 and alpha = 0.05.

t(critical) = 1.658

Therefore, | t | > t(critical) and hence , we reject the null hypothesis at alpha = 0.05.

We conclude that in average students exercise less than 15 hours per month.