Question

Researchers were interested in comparing the long-term psychological effects of being on a high-carbohydrate, low-fat (LF) diet versus a high-fat, low-carbohydrate (LC) diet. A total of 106 overweight and obese participants were randomly assigned to one of these two energy-restricted diets. At 52 weeks, 32 LC dieters and 33 LF dieters remained. Mood was assessed using a total mood disturbance score (TMDS), where a lower score is associated with a less negative mood. A summary of these results follows: Group n ¯ x s LC 32 47.3 28.3 LF 33 19.3 25.8 Is there a difference in the TMDS at Week 52? Test the null hypothesis that the dieters’ average mood in the two groups is the same. Use a significance level of 0.05. Question 7 (2 points total) In the previous question, the data was analyzed using the two-sample t test that does not assume equal standard deviations. Repeat the analysis, but conduct a pooled two-sample t test that does assume equal standard deviations.

Answer 1

Solution:

Here, we have to use two sample t test for the difference between two population means assuming equal population variances or standard deviations.

Null hypothesis: H₀: The dieters’ average mood in the two groups is the same.

Alternative hypothesis: H_a: The dieters’ average mood in the two groups is not the same.

H₀: µ₁ = µ₂ versus H_a: µ₁ ≠ µ₂

We are given

Level of significance = α = 0.05

The test statistic formula for this test is given as below:

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

From given data, we have

X1bar = 47.3

X2bar = 19.3

S1 = 28.3

S2 = 25.8

n1 = 32

n2 = 33

df = n1 + n2 – 2 = 32 + 33 – 2 = 63

α = 0.05

Critical values = - 1.9983 and 1.9983

(by using t-table)

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

S_p² = [(32 – 1)* 28.3^2 + (33 – 1)* 25.8^2]/(32 + 33 – 2)

S_p² = 732.1916

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

t = (47.3 – 19.3) / sqrt[732.1916*((1/32)+(1/33))]

t = 28/6.7133

t = 4.1708

P-value = 0.0001

(by using t-table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

There is insufficient evidence to conclude that the dieters’ average mood in the two groups is the same.