Question

You wish to test the following claim (H1) at a significance level of α=0.005.

      Ho:μ1=μ2
      H1:μ1>μ2

You obtain a sample of size n1=117 with a mean of M1=62.8 and a standard deviation of SD1=10.2 from the first population. You obtain a sample of size n2=114 with a mean of M2=57.5 and a standard deviation of SD2=12.8mfrom the second population.

What is the critical value for this test? (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 greater than the second population mean.
• There is not sufficient evidence to warrant rejection of the claim that the first population mean is greater than the second population mean.
• The sample data support the claim that the first population mean is greater than the second population mean.
• There is not sufficient sample evidence to support the claim that the first population mean is greater than the second population mean.

Answer 1

   Ho:μ1=μ2
H1:μ1>μ2

Level of Significance(l.o.s.) :  $\alpha$ = 0.005

Decision Criteria : Reject Ho at 0.5% l.o.s. if t cal > t tab,
where critical value = t tab = t (0.005, n1+n2-2) = 2.5975

Calculation : n1=117, n2 = 114,  $\bar{x}_{1}$ = 62.8,   = 57.5, s1 = 10.2, s2 = 12.8

22

s =  $\sqrt{\frac{(n_{1}-1)*s_{1}^{2}+(n_{2}-1)*s_{2}^{2}}{n_{1}+n_{2}-2}}$ = 11.5563

t cal =   $\frac{\bar{x}_{1}-\bar{x}_{2}}{s\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}$ =  $\frac{62.8-57.5}{11.5563*\sqrt{\frac{1}{117}+\frac{1}{114}}}$ = 3.4850

Conclusion : The test statistic is in the critical region.
Since t cal > t tab, we reject Ho at 0.5% l.o.s.

As such, the final conclusion is that :
There is not sufficient evidence to warrant rejection of the claim that the first population mean is greater than the second population mean.

Hope this answers your query!