Question

You wish to test the following claim (HaHa) at a significance level of α=0.05α=0.05.

      Ho:μ=72.2Ho:μ=72.2
      Ha:μ<72.2Ha:μ<72.2

You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n=31n=31 with mean M=67.7M=67.7 and a standard deviation of SD=15.6SD=15.6.

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic =

What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =

Answer 1

Given ,

Sample size = 31

Sample mean = = 67.7

Sample standard deviation = 15.6

Hypothesis :

Left tailed test.

Test statistic :

Test statistic = t = -1.606

P-value :

P-value for this left tailed test is ,

P-value = P( t < test statistic ) = P( t < -1.606 )

Using Excel function ,   =T.DIST( t , df , 1 )   , df = n - 1 = 31- 1 = 30

P( t < -1.606 ) = T.DIST( -1.606 , 30 , 1 ) = 0.0594

P-value = 0.0594

Decision about null hypothesis :

Rule : Reject null hypothesis if p-value less than significance level

Here = 0.05

It is observed that p-value ( 0.0594 ) is greater than = 0.05

So fail to reject null hypothesis.

Conclusion :

There is not sufficient evidence to support Ha : < 72.2