Question

A researcher asks if people score higher or lower on a questionnaire measuring their emotional wellbeing when they are exposed to much sunshine compared to when they’re exposed to little sunshine. A sample of people is measured under both levels of sunshine and produced the well-being scores below:

a. What are my null and alternative hypotheses?
b. Calculate the appropriate z or t test statistics.
c. What can you conclude about the hypotheses and write an appropriate concluding
statement?
d. Do you run the risk of committing a Type 1 or Type 2 error? Why?

Low High
14 18
13 12
17 20
15 19
18 22
17 19
14 19
16 16

Answer 1

The paired t-test was used to compare emotional well-being scores between people exposed to little and much sunshine. The t-test statistic was 3.89, leading to rejection of the null hypothesis and a conclusion of a significant difference.

a. Null hypothesis (H0): There is no difference in emotional well-being scores between people exposed to little sunshine and people exposed to much sunshine.

Alternative hypothesis (H1): People score differently on the emotional well-being questionnaire when exposed to much sunshine compared to when they're exposed to little sunshine.

b. To test the hypotheses, we can use a paired t-test since the same group of people is measured under both levels of sunshine. The formula for the t-test statistic for paired samples is:

t = (mean of the differences) / (standard error of the differences)

First, calculate the differences between the high and low sunshine scores:

Differences: (18-14), (12-13), (20-17), (19-15), (22-18), (19-17), (19-14), (16-16)

Differences: 4, -1, 3, 4, 4, 2, 5, 0

Next, calculate the mean of the differences:

Mean = (4-1+3+4+4+2+5+0) / 8 = 21 / 8 = 2.625

Then, calculate the standard deviation of the differences:

s = √((Σ(x-mean)^2) / (n-1))

s = √((4-2.625)^2 + (-1-2.625)^2 + (3-2.625)^2 + (4-2.625)^2 + (4-2.625)^2 + (2-2.625)^2 + (5-2.625)^2 + (0-2.625)^2) / 7)

s ≈ √(1.953 + 12.953 + 0.078 + 1.953 + 1.953 + 0.422 + 6.328 + 6.891) / 7

s ≈ √(25.578) / 7

s ≈ √3.654

s ≈ 1.91

Now, calculate the standard error of the differences:

SE = s/√n

SE = 1.91/√8

SE ≈ 0.675

Finally, calculate the t-test statistic:

t = 2.625 / 0.675

t ≈ 3.89

c. The t-test statistic calculated is 3.89. This value falls into the rejection region for a significance level of α = 0.05 with 7 degrees of freedom. Therefore, we reject the null hypothesis and conclude that there is a significant difference in emotional well-being scores between people exposed to little sunshine and people exposed to much sunshine.

d. By rejecting the null hypothesis when it is actually true, we would commit a Type 1 error in this case. This means that we would conclude that there is a difference in emotional well-being scores when there isn't one in reality.