For a random sample of 10 observations the sample mean was 1

For a random sample of 10 observations, the sample mean was 10 and the sample standard deviation 3.20. Using the .01 significance level:

Solution

(a) The degree of freedom =n-1=10-1=9

Given a=0.01, the critical value is t(0.01, df=9) =2.821 (from student t table)

Reject H₀ if t >2.821

---------------------------------------------------------------------------------------------------------------------------------

(b)test statistic is

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

=(10-8)/(3.2/sqrt(10))

=1.98

---------------------------------------------------------------------------------------------------------------------------------

(c)Cannot reject H₀. The mean is not greater than 8.