Information Use the following scenario to answer questions 18-22: A physician ha

ID: 3061783 • Letter: I

Question

Information

Use the following scenario to answer questions 18-22: A physician has developed three different weight loss diets. As an experiment, she randomly selected 15 patients and assigned 5 to each diet. After 3 weeks, the mean amount of weight loss in each diet plan was computed. Can she conclude that there is a difference in the mean amount of weight loss among the three diets? Use alpha = 0.05. Based on the sample data, the following quantities are computed: SSTreatment = 26.13 SSTotal= 37.73.

Question 18 (1 point)

Write the null hypothesis to test whether there is a difference between the diets the doctor is investigating.

H0: at least one of the population means is different

H0: 1=2=3

s1=s2=s3

H0: p1=p2=p3

Question 19 (1 point)

How many degrees of freedom does the sum of squares for error have?

Question 20 (1 point)

What is the mean sum of squares for treatment?

3.86

5.8

8.71

0.96

13.065

Question 21 (1 point)

What is the calculated value of the test statistic for this test?

2.66

11.26

13.52

4.74

2.25

Question 22 (1 point)

What is the MANAGERIAL conclusion based on this hypothesis?

Reject the null hypothesis.

Do not reject the null hypothesis.

At least one diet plan has a different mean weight loss amount compared to the others.

The mean amount of weight loss among the three diets is the same.

The two diet plans have the same mean amounts of weight loss.

Explanation / Answer

Question 18:

Write the null hypothesis to test whether there is a difference between the diets the doctor is investigating.

Sol: Here we are supposed to test whether there is significant difference between the three different weight loss diets doctor is investigating.

Here the null hypothesis is there is no difference between population means for three weight loss diets which can be stated mathematically as follows:

H0: 1=2=3

To answer other questions, we need to create ANOVA table from given information.

n=number of observations=15

m=number of treatments=3

The ANOVA table is given as follows:

Source

Degrees of freedom (DF)

Sum of squares (SS)

Mean sum of squares (MSS)

[SS/DF]

F-value

Treatment

m-1=2

26.13

26.13/2=13.065

13.065/0.966=13.5155

Error

n-m=12

SSTotal-SSTreatment=37.73-26.13=11.6

37.73/12=0.966

Total

n-1=14

37.73

Question 19:

Sol: How many degrees of freedom does the sum of squares for error have?

As per above table, there are 12 degrees of freedom for error.

Question 20:

Sol: What is the mean sum of squares for treatment?

Mean sum of squares for treatment is 13.065

Question 21:

Sol: What is the calculated value of the test statistic for this test?

The value is test statistic is F value in above table i.e. 13.51

Question 22:

Sol: What is the MANAGERIAL conclusion based on this hypothesis?

Here F statistic follows F distribution with (2, 12) degrees of freedom.

The F probability table gives critical value as 3.88

Observe that the calculated value is greater than critical value, hence we can reject the null hypothesis.

Hence we can conclude that: