Question

The human resources department of a consulting firm gives a standard creativity test to a randomly selected group of new hires every year. This year, 70 new hires took the test and scored a mean of 112.5 points with a standard deviation of 15.2. Last year, 55 new hires took the test and scored a mean of 117.5 points with a standard deviation of 18. Assume that the population standard deviations of the test scores of all new hires in the current year and the test scores of all new hires last year can be estimated by the sample standard deviations, as the samples used were quite large. Construct a 95% confidence interval for −μ1μ2, the difference between the mean test score μ1 of new hires from the current year and the mean test score μ2 of new hires from last year. Then complete the table below.

Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places.

What is the lower limit of the 95% confidence interval? What is the upper limit of the 95% confidence interval?

Answer 1

Here we have given that

n1=number of hires took the test in 1st sample =70

=first sample mean = 112.5

S1=first sample standard deviation =15.2

n2=number of hires took the test in 2nd sample = 55

=Second sample mean = 117.5

S2=second sample standard deviation =18

Now we want to find

95% confidence interval for the difference between the mean test score μ1 of new hires from the current year and the mean test score μ2 of new hires from last year i.e. ().

For finding this CI 1^st we want to check the two population variance equal or not.

Formula is as follows

Claim: To check whether the two population variance equal or not.

The hypothesis is

v/s

Test statistics is

F-statistics=

=

= 0.71

c=confidence level =0.95

=level of significance=1-c=1-0.95 = 0.05

Degrees of freedom 1=n1-1=70-1=69

Degrees of freedom 2= n2-1= 55-1=54

F-critical = 1.54 Using Excel =FINV(probability=0.05, D.F1=69, D.F2=54)

Decision:

Here F-tabulated < F-critical

Then we fail to reject Null hypothesis that the population variances are equal.

We want to find he 95% confidence interval for difference in the two population means ().

Formula is as follows

Where

Sp=

Now we find the Spooled

Sp =

     =

     = 16.488

And

Now, we find the t-critical value

c=confidence level =0.95

=level of significance=1-c=1-0.95 = 0.05

Degrees of freedom 1= n1+n2-2 =70+55-2 =123

we get

t-critical=1.979 Using excel=TINV(prob=0.05, d.f=123)

We get the 95% confidence interval is

Lower limit= -10.88

Upper limit= 0.88

Interpretation:

This confidence interval shows the we are 95% confidence that the difference between the mean test score μ1 of new hires from the current year and the mean test score μ2 of new hires from last year i.e. () will falls within this interval.