Question

You wish to test the following claim (HaHa) at a significance level of α=0.01α=0.01.

      Ho:μ1=μ2Ho:μ1=μ2
      Ha:μ1≠μ2Ha:μ1≠μ2

You believe both populations are normally distributed, but you do not know the standard deviations for either. However, you also have no reason to believe the variances of the two populations are not equal. You obtain a sample of size n1=15n1=15 with a mean of M1=76.2M1=76.2 and a standard deviation of SD1=12.6SD1=12.6 from the first population. You obtain a sample of size n2=18n2=18 with a mean of M2=73.1M2=73.1 and a standard deviation of SD2=12.5SD2=12.5 from the second population.

What is the critical value for this test? For this calculation, use the conservative under-estimate for the degrees of freedom as mentioned in the textbook. (Report answer accurate to three decimal places.)
critical value = ±±

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

The test statistic is...

• in the critical region
• not in the critical region

This test statistic leads to a decision to...

• reject the null
• accept the null
• fail to reject the null

As such, the final conclusion is that...

• There is sufficient evidence to warrant rejection of the claim that the first population mean is not equal to the second population mean.
• There is not sufficient evidence to warrant rejection of the claim that the first population mean is not equal to the second population mean.
• The sample data support the claim that the first population mean is not equal to the second population mean.
• There is not sufficient sample evidence to support the claim that the first population mean is not equal to the second population mean.

Answer 1

Here we have : n1 = 15, M1= 76.2, s₁=12.6, n2 = 18, M2 = 73.1, s₂= 12.5

The hypothesis are

Ho:μ1=μ2 v/s  Ha:μ1≠μ2

The critical values are :

conservative df = smaller ( n1-1,n2-1 ) = smaller ( 15-1, 18-1 ) = 14, α=0.01

Critical values = ------ ( using excel formula " =t.inv.2t(0.01,14)" )

Reject null hypothesis if , test statistic lies in the critical region. i.e.

t > 2.977 or t < -2.977

The test statistic is,

  

= 0.706

Here calculated value of t does not lie in the critical value . Hence reject null hypothesis.

Conclusion : There is not sufficient sample evidence to support the claim that the first population mean is not equal to the second population mean.