Question

1. To evaluate the effect of a treatment, a sample is obtained from a population with an estimated mean of μ = 30, and the treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be 32.5, with a standard deviation of s = 3. (5 points)
   1. Which statistical test is appropriate for this problem: one-sample t-test, repeated measures t-test, or independent samples t-test?

2. If the sample consists of n = 16 individuals, are the data sufficient to conclude that the treatment had a significant effect using a two-tailed test with α = .05?

3. Calculate Cohen’s d.

4. Provide an interpretation of the analysis in APA format.

Answer 1

To evaluate the effect of a treatment, a sample is obtained from a population with an estimated mean of μ = 30,.

After treatment, the sample mean is found to be 32.5, with a standard deviation of s = 3.

a. Which statistical test is appropriate for this problem: one-sample t-test, repeated measures t-test, or independent samples t-test?

- Here we are given only one sample of After treatment ,

Hence one-sample t-test is appropriate for this problem.

b. If the sample consists of n = 16 individuals, are the data sufficient to conclude that the treatment had a significant effect using a two-tailed test with α = .05?

- The parametric test called t-test is useful for testing those samples whose size is less than 30. A small sample is generally regarded as one of size n<30. A t-test is necessary for small samples because their distributions are not normal. So here we can assume data to be sufficient .

c . Calculate Cohen’s d.

Now Cohen’s d is given by

d = = = 0.83333

Now , we have to test

H0 : μ = 30           { treatment had a significant effect }

H1 : μ 30        { treatment do not have a significant effect }

Cohen’s d for one-sample t-test statistics is

Thus   t =

where is sample mean , s is sample standard devitation , and n = 16 thus n^1/2 = 4

Hence

t = = = 3.333333

We reject null hypothesis is calculated t -value is greater than

where is t-distributed with n-1 = 15 degree of freedom and = 0.05

It can be computed from statistical book or more accurately from any software like R,Excel

From R

> qt(1-0.05/2,df=15)
[1] 2.13145

thus = 2.13145

Now calculated absolute of t -value | t | = 3.333333 is greater than 2.13145

i.e | t | = 3.333333 > 2.13145

Hence   | t | >

So we reject null hypothesis at 5% of significance level .

d. Provide an interpretation of the analysis in APA format.

Since we have rejected null hypothesis H0 at 5% of significance level ,

We conclude that treatment may not have a significant effect