Question

1. Researchers have noted a decline in cognitive functioning as people age (Bartus, 1990). However, the results from other research suggests that the antioxidants in foods such as blueberries can reduce and even reverse these age-related declines, at least in laboratory rats (Joseph et al., 1999). Based on these results, one might theorize that the same antioxidants might also benefit elderly humans. Suppose a researcher is interested in testing this theory. The researcher obtains a sample of n = 16 adults who are older than 65, and gives each participant a daily dose of a blueberry supplement that is very high in antioxidants. After taking the supplement for 6 months, the participants are given a standardized cognitive skills test and produce M = 54. For the general population of elderly adults, scores on the test average μ = 50 and form a normal distribution with σ = 12.

Do the data from this sample provide evidence that the blueberry supplement increases cognitive skills among elderly people? Using a one-tail test with α = .05.

a. State the null hypothesis in words and in a statistical form.

b. State the alternative hypothesis in words and a statistical form.

c. Compute the appropriate statistic to test the hypotheses. Sketch the distribution with the standard error and locate the critical region with the critical value

d. State your statistical decision. (.5)

e. Compute Cohen’s d to measure the size of the effect. Interpret what this effect size

really means in this context.

f. What is your conclusion? (don't forget the statistical information).

g. Given your statistical decision (in part d), what type of decision error could you have made

Answer 1

M= 54, n=16, μ = 50, σ = 12, α = .05

a) State the null hypothesis in words and in a statistical form.

H0: μ1 = μ2 , Mean scores of the test before and after giving blueberry supplement are the same

b) State the alternative hypothesis in words and a statistical form.

H1: μ1 < μ2 , Mean scores of the test before giving blueberry supplement are less than scores of the test after giving blueberry supplement

c) Compute the appropriate statistic to test the hypotheses. Sketch the distribution with the standard error and locate the critical region with the critical value

Test statistic; z-test: z = (M-μ)/(σ/n^0.5) =(54-50)/(12/16^0.5) = 1.33

Standard error, SE = σ/n^0.5 = 12/16^0.5 = 3

The shaded region in the figure is the critical region.

d) State your statistical decision.

Tabulated z = zα/2 = z0.025= 1.96

Since calculated z< tabulated z, we do not reject null hypothesis.

e) Compute Cohen’s d to measure the size of the effect. Interpret what this effect size really means in this context.

d = |(M-μ)/σ| = |(54-50)/16| = 0.25

Thus, difference between test scores before and after giving blueberry supplement is relatively small (0.25 standard deviation).

f) What is your conclusion?

H0: μ1 = μ2 , Mean scores of the test before and after giving blueberry supplement are the same

H1: μ1 < μ2 , Mean scores of the test before giving blueberry supplement are less than scores of the test after giving blueberry supplement

Test statistic; z-test: z = (M-μ)/(σ/n^0.5) =(54-50)/(12/16^0.5) = 1.33

Tabulated z = zα/2 = z0.025= 1.96

Since calculated z< tabulated z, we do not reject null hypothesis and conclude that mean scores of the test before and after giving blueberry supplement are the same.

g) Given your statistical decision (in part d), what type of decision error could you have made

A type 1 error (α) of rejecting a null hypothesis that is true could have been made.