Question

A weight-lifting coach wanted to know whether weight-lifters can change their strength by taking a certain supplement. To answer this question, the coach randomly selected 8 athletes and gave them a strength test using a bench press. Thirty days later, after regular training using the supplement, they were tested again. The results were listed below. A test was conducted to determine whether weight-lifters can change their strength by taking a certain supplement. Assume the populations are normally distributed.

Athlete 1 2 3 4 5 6 7 8 Mean SD

Before 215 240 188 212 275 260 225 200 226.88 29.72

After 225 245 188 210 282 275 230 195 231.25 34.55

Difference -10 -5 0 2 -7 -15 -5 5 -4.38 6.55

1. What is the null hypothesis for this test?

2. What is the value of the standardized test statistic?

3. The p-value was 0.05. Suppose this test was conducted at α = 0.01. What can you conclude?

a. There is insufficient evidence to conclude that weight-lifters can change their strength by taking a certain supplement.

b. There is sufficient evidence to conclude that weight-lifters can change their strength by taking a certain supplement.

c. None of the above; this test is invalid.

4. Suppose the coach had incorrectly performed a two-sample t-test. In comparison to the correct analysis, which of the following regarding the test statistic would be true?

a. The test statistic is closer to 0

b. The test statistic is farther from 0

c. The test statistic does not change.

Answer 1

Solution:

Here, we have to use paired t test.

The null and alternative hypotheses for this test are given as below:

1. What is the null hypothesis for this test?

Null hypothesis: H₀: The weight-lifters cannot change their strength by taking a certain supplement.

Alternative hypothesis: H_a: The weight-lifters can change their strength by taking a certain supplement.

H₀: µ_d = 0 versus H_a: µ_d ≠ 0

This is a two tailed test.

2. What is the value of the standardized test statistic?

Test statistic for paired t test is given as below:

t = (Dbar - µ_d)/[S_d/sqrt(n)]

From given data, we have

Dbar = -4.38

S_d = 6.55

n = 8

df = n – 1 = 7

α = 0.01

t = (Dbar - µ_d)/[S_d/sqrt(n)]

t = (-4.38 – 0)/[6.55/sqrt(8)]

t = -1.8914

3. The p-value was 0.05. Suppose this test was conducted at α = 0.01. What can you conclude?

P-value = 0.05 > α = 0.01, so we do not reject the null hypothesis

a. There is insufficient evidence to conclude that weight-lifters can change their strength by taking a certain supplement.

4. Suppose the coach had incorrectly performed a two-sample t-test. In comparison to the correct analysis, which of the following regarding the test statistic would be true?

a. The test statistic is closer to 0

...because difference between two sample means will be close to zero