A researcher was interested in assessing the effectiveness of the Statistics Diet as

compared to a regular low calorie diet on weight loss. In the study, obese participants were
randomly assigned to one of two groups: (1) the Statistics Diet, which required participants
to calculate the mean number of calories in each food that they ate at every meal or (2) the
Regular Low Calorie Diet, which had participants consume only 1000 calories a day. The
dependent/response measure collected was the number of pounds lost at the end of the
first week of the diet. The following table contains the results of the study.
#ofParticipnts Mean Stndard Dev
Stats Diet 6 2.3 0.55
Reg Low Cal Diet 6 1.23 0.5
What is the value of the test statistic?
a. 3.53
b. -3.53
c. 11.62
d. -11.62

To find the value of the test statistic, we need to perform an independent samples t-test.

The test statistic formula for an independent samples t-test is:
t = (mean1 - mean2) / √((s^21 / n1) + (s^2/2 / n2))

Given the information from the table:
mean1 = 2.3
mean2 = 1.23
s1 = 0.55
s2 = 0.5
n1 = 6
n2 = 6

Plugging in the values into the formula, we get:
t = (2.3 - 1.23) / √((0.55^2 / 6) + (0.5^2 / 6))

Calculating the values inside the square root:
t = (2.3 - 1.23) / √((0.3025 / 6) + (0.25 / 6))

Simplifying further:
t = (2.3 - 1.23) / √(0.05042 + 0.04167)

Adding the values inside the square root:
t = (2.3 - 1.23) / √(0.09209)

Calculating the value inside the square root:
t = (2.3 - 1.23) / 0.3034

Performing the subtraction:
t = 1.07 / 0.3034

Calculating the division:
t = 3.5258

Therefore, the value of the test statistic is approximately 3.53.

The correct answer is:
a. 3.53

To find the value of the test statistic, we need to conduct a statistical test to compare the means of the two groups.

In this case, we have two groups: the Statistics Diet group and the Regular Low Calorie Diet group. The null hypothesis (H0) is that there is no difference in weight loss between the two diets, and the alternative hypothesis (Ha) is that there is a difference.

The appropriate statistical test for comparing two means is the independent samples t-test. The test statistic for the independent samples t-test is calculated using the formula:

t = (mean1 - mean2) / sqrt((s1^2 / n1) + (s2^2 / n2))

Where:
mean1 and mean2 are the means of the two groups
s1^2 and s2^2 are the variances of the two groups
n1 and n2 are the sizes of the two groups

Using the values given in the table, we can substitute them into the formula to calculate the test statistic:

mean1 = 2.3
mean2 = 1.23
s1 = 0.55
s2 = 0.5
n1 = 6
n2 = 6

t = (2.3 - 1.23) / sqrt((0.55^2 / 6) + (0.5^2 / 6))
= 1.07 / sqrt(0.050511 + 0.046667)
= 1.07 / sqrt(0.097178)
≈ 1.07 / 0.31171
≈ 3.43

Therefore, the value of the test statistic is approximately 3.43.

The closest option given in the answer choices is 3.53, so the correct answer is a. 3.53.