The national average of times a person checks their email is

The national average of times a person checks their email is 32.7 times per month. A random sample of 20 people yielded a mean of 45.4 email checks per month with a standard deviation of 5.3. At the 0.05 level of significance can it be concluded that this differs from the national average?

Solution

Let mu be the population mean

Ho: mu=32.7(i.e. null hypothesis)

Ha: mu not equal to 32.7 (i.e. alternative hypothesis)

The test statistic is

t=(xbar-mu)/(s/vn)

=(32.7-45.4)/(5.3/sqrt(20))

=-10.72

The degree of freedom =n-1=20-1=19

It is a two-tailed test.

Given a=0.05, the critical values are t(0.025,df=19) =-2.09 or 2.09 (from student t table)

Since t=-10.72 is less than -2.09, we reject the null hypothesis.

So we can conclude that this differs from the national average