Question

	2008	2007
Sample Size	36	36
Sample Mean	75.0000	70.7500
Sample Standard Deviation	17.2891	8.7843
df	51
Confidence Interval (in terms of 2008 - 2007)
Confidence Coefficient	0.80
Lower Limit	0.0535
Upper Limit	8.4465
Hypothesis Test (in terms of 2008 - 2007)
Hypothesized Value	0
Test Statistic
p-value (Lower Tail)
p-value (Upper Tail)
p-value (Two Tail)

	2008-2007
Sample Size	36
Sample Mean	4.2500
Sample Standard Deviation	20.2969
Confidence Interval (in terms of 2008 - 2007)
Confidence Coefficient	0.80
Lower Limit	-0.1687
Upper Limit	8.6687
Hypothesis Test (in terms of 2008 - 2007)
Hypothesized Value	0
Test Statistic
p-value (Lower Tail)
p-value (Upper Tail)
p-value (Two Tail)

During 2008, JCPenney took steps to improve customer service in order to increase their customer satisfaction score. A random sample of 36 JCPenney customers in the 4th quarter of 2007 and a random sample of 36 JCPenney customers in the 4th quarter of 2008 were surveyed and their customer satisfaction score entered into Excel. Can JCPenney conclude the average 2008 customer satisfaction score is greater than the average 2007 customer satisfaction score at α=.005? Based on this paragraph of text, use the correct excel output above to answer the following question.

What would the P-value be for H_a: μ₂₀₀₇ - μ₂₀₀₈ < 0?

	a.	.1 < P-value < .5
	b.	.95 < P-value < .975
	c.	None of the answers is correct
	d.	.90 < P-value < .95
	e.	.05 < P-value < .1

Answer 1

Here we are comparing different sets of customers in both years. So the samples are independent of each other. We need to test for the difference of the means. Since there is a larger difference between the Sample SDs we will assume they have unequal population variances. So we will use indepenent t-test for difference of population means.

	2007 (1)	2008 (2)
n	36	36
mean	70.75	75
sd	8.7843	17.2891

In the question it is in terms of μ2007 - μ2008 so we will find everything the same way.

Using the following formulas we will find the test stat and df

Test Stat = -1.3149

v = n - 1 there v1 = v2 = 35

df = 51.94 = 52

Since we checking if μ2007 - μ2008 < 0, it is one tailed test

p-value = P( > |Test Stat|)

= P( > 1.31)

p-value = 0.0972

e.	.05 < P-value < .1

Since p-value > 0.005

We do not reject the null hypothesis at 0.005 level.

Please give it a thumps up if the solution helped you.