# Chapter 9: Paired- and Independent-Samples t-Test

Answers for Data Skill Challenges for every chapter in the book can be found to check your performance and widen your understanding.

1)      Although ‘before’ and ‘after’ are part of the problem, the data are not paired – there is no link between any particular day and any other day. Instead, we want to compare ‘customers who shopped before the change’ and ‘customers who shopped after the change’, which may be considered two different statistical populations to be compared with each other. That means we need to calculate independent-samples t.

RQ: Do customers who shop after the policy change make more purchases than those who shopped before the policy change?

H0: μ1μ2 ≤ 0

H1: μ1μ2 > 0

where μ1 represents those shopping after the change and μ2 represents those shopping before the change

α = .05

tcrit(18) = +1.734

By hand: t(18) = 1.77, p <.05

SPSS/Excel: t(18) = 1.77, p = .046

d = 0.78

Conclusion: Reject the null and accept the alternative. The difference is statistically significant. Customers who shop after the policy change make more purchases than those who shopped before the policy change. The difference between those shopping before and after the policy change is a medium effect.

2)      In this problem, there is a link between the two pieces of data. Each store is represented twice – before change and after change. That means the data are paired, and we need to calculate paired-samples t.

RQ: Is there a difference in the amount of purchases made after the policy change compared to before the policy change?

H0: μD = 0

H1: μD ≠ 0

where μD = μTime2 μTime1

α = .05

t(9)crit = ±2.262

CI95 = [–2.72, 15.32]

By hand: t(9) = 1.58, p > .05

SPSS/Excel: t(9) = 1.58, p = .149

Conclusion: Retain the null. The difference is not statistically significant. There is insufficient evidence to conclude that there is a difference in the amount of purchases made after the policy change compared to before the policy change. If we assume this difference to represent the population, we would expect 95% of drawn sample mean differences to fall between –2.72 and 15.32.

3)      Since we want to compare ‘customers who shop in stores in Region A’ and ‘customers who shop in stores in Region B’, which may be considered two different statistical populations to be compared with each other, we need to calculate independent-samples t.

RQ: Do customers who shop after the policy change in stores in Region A report higher satisfaction than customers who shop in Region B?

H0: μ1μ2 ≤ 0

H1: μ1μ2 > 0

where μ1 represents those shopping in Region A and μ2 represents those shopping in Region B

α = .05

tcrit(18) = +1.734

By hand: t(18) = 1.48, p >.05

SPSS/Excel: t(18) = 1.48, p = .078

Conclusion: Retain the null. The difference is not statistically significant. There is insufficient evidence to conclude that customers who shop after the policy change in stores in Region A report higher satisfaction than those who shop in Region B.

4)      In this problem, there is a link between the two pieces of data. Each employee is represented twice – before change and after change. That means the data are paired, and we need to calculate paired-samples t.

RQ: Do employees report different satisfaction scores after the change compared to before the change?

H0: μD = 0

H1: μD ≠ 0

where μD = μTime2 μTime1

α = .05

t(9)crit = ±2.262

CI95 = [–3.24, –.76]

By hand: t(9) = –4.47, p < .05

SPSS/Excel: t(9) = –4.47, p = .011

Conclusion: Reject the null and accept the alternative. The difference is statistically significant. There is sufficient evidence to conclude that there is a difference in employee satisfaction before and after the policy change. If we assume this difference to represent the population, we would expect 95% of drawn sample mean differences to fall between –3.24 and –.76.