Question

A sample is selected from a population with µ= 50. After a treatment is administered to the individuals in the sample, the mean is found to be M= 55 and the variance is s²= 64.

          a. For a sample of n = 4 scores, conduct a single sample t-test to evaluate the  significance of the treatment effect and calculate Cohen’s d to measure the size of the treatment effect. Use a two-tailed test with α = .05.Show the sampling distribution.(2pts)

          b. For a sample of n = 16 scores, conduct a single sample t-test to evaluate the significance of the treatment effect and calculate Cohen’s d to measure the size of the treatment effect. Use a two-tailed test with α = .05. Show the sampling distribution. (2 pts)

          c. Using symbols, write up your results. Describe how increasing the size of the sample affects the likelihood of rejecting the null hypothesis and the measure of effect size. (1pt)

Looking for the answer to C, but answers for a and b could be appreciate.

Answer 1

(a) The hypothesis being tested is:

H₀: µ = 50

H_a: µ ≠ 50

The test statistic, t = (x - µ)/s/√n

t = (55 - 50)/8/√4

t = 1.25

The p-value is 0.2999.

Since the p-value (0.2999) is greater than the significance level (0.05), we fail to reject the null hypothesis.

Therefore, we cannot conclude that µ ≠ 50.

Cohen's d = (55 - 50)/8 = 0.625

(b) The hypothesis being tested is:

H₀: µ = 50

H_a: µ ≠ 50

The test statistic, t = (x - µ)/s/√n

t = (55 - 50)/8/√16

t = 2.5

The p-value is 0.0245.

Since the p-value (0.0245) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that µ ≠ 50.

Cohen's d = (55 - 50)/8 = 0.625

(c) Increasing the sample size decreases the probability because as the sample size increases, the standard error decreases.

The effect size would still be the same since it is independent of the sample size.