Question

A random sample of 50 measurements resulted in a sample mean of 62 with a sample standard deviation 8. It is claimed that the true population mean is at least 64.

(a) Is there sufficient evidence to refute the claim at the 2% level of significance?

(b) What is the p-value?

(c) What is the smallest value of α for which the claim will be rejected?

Answer 1

Solution

Given:

The sample mean is  = 62 and the Sample standard deviation is s = 8

The sample size is n = 50.

Null and Alternative Hypotheses:

                                       

This corresponds to a two-tail test, for which a t-test for one mean, with unknown population standard deviation will be used.

Test statistic:

Formula:

                                   

The number of degrees of freedom are df = n-1 = 50-1=49

P-value approach:
Given significance level = α = 0.02 and df = 49
P(T<-1.768) = 0.0417 …Using excel formula: =TDIST(1.768,49,1)

Since it is observed that P-value = 0.0417 > 0.02
It is then concluded that the Null Hypothesis is not rejected.

Therefore is sufficient evidence to claim that true population mean is atleast 64 at 0.02 level of significance.

b) The number of degrees of freedom are df = n-1 = 50-1=49

P-value approach:
Given significance level = α = 0.02 and df = 49
P(T<-1.768) = 0.0417 …Using excel formula: =TDIST(1.768,49,1)

c) If we take significance level = α = 0.05 then,

P-value = 0.0417 < 0.05
It is then concluded that the Null Hypothesis is rejected.