Question

In: Statistics and Probability

Problem 1: A psychologist is interested in a diet plan that is said to work better...

Problem 1: A psychologist is interested in a diet plan that is said to work better than any other diet. To test this the psychologist randomly assigns 32 friends to one of four diet plans: Life of Herbs (LOH), E-Mail-A-Meal (EAM), Weight-Lookers (WLK), and Low-Carbs (LOC). The numbers represent weight loss in pounds for three months for each friend.  
LOH-14,19,12,9,14,15,18,8
EAM-11,15,9,19,12,10,20,8
WLK-10,5,6,11,4,5,4,3
LOC-25,18,24,29,17,20,23,27

The sum of squares between groups (SSB) for this problem is

a) 1150.75

b) 436.75

c) 1587.50

d) 24.591

The sum of squares within groups (SSW) for Problem 1 is

a) 1150.75

b) 436.75

c) 1587.50

d) 24.591

The sum of squares total (SST) for Problem 1 is

a) 1150.75

b) 436.75

c) 1587.50

d) 24.591

The F value for Problem 1 is

a) 1150.75

b) 436.75

c) 1587.50

d) 24.591

the degrees of freedom for the numerator (df1) is

a) 3

b) 28

c) 31

d) 1

the degrees of freedom for the denominator (df2) is

a) 3

b) 28

c) 31

d) 1

The mean squares within groups (MSW) is

a) 383.583

b) 436.750

c) 15.598

d) 24.591

The null hypothesis for Problem 1 should be

a) rejected

b) not rejected

c) retained

d) none of the above

Which of the following can be concluded about the Problem 1 experiment?

a) There are differences in weight loss among the groups.

b) There are no differences in weight loss among the groups.

c) One group was superior to the other groups in weight loss.

d) Two of the groups were better than the others in weight loss.

Solutions

Expert Solution

The sum of squares between groups (SSB) for this problem is

Ans:

a) 1150.75

The sum of squares within groups (SSW) for Problem 1 is

Ans:

b) 436.75

The sum of squares total (SST) for Problem 1 is

Ans:

c) 1587.50

The F value for Problem 1 is

Ans:

d) 24.591

the degrees of freedom for the numerator (df1) is

Ans:

a) 3

the degrees of freedom for the denominator (df2) is

Ans:

b) 28

The mean squares within groups (MSW) is

Ans:

a) 383.583

The null hypothesis for Problem 1 should be

Ans:

a) rejected

Which of the following can be concluded about the Problem 1 experiment?

Ans:

a) There are differences in weight loss among the groups.

#######################Explanation#################

Let and be the mean weight loss due to four diet plans: Life of Herbs (LOH), E-Mail-A-Meal (EAM), Weight-Lookers (WLK), and Low-Carbs (LOC) respectively

The null hypothesis for an ANOVA always assumes the population means are equal.

Hence, the null hypothesis as:

i.e There are no differences in weight loss among the groups.

and the alternative hypothesis is given by

i.e There are differences in weight loss among the groups.

The given dataset is

LOH EAM WLK LOC
14 11 10 25
19 15 5 18
12 9 6 24
9 19 11 29
14 12 4 17
15 10 5 20
18 20 4 23
8 8 3 27

Let Xij be the observations.

i = 1,2 ...r. r is total number of rows

j= 1,2,...t t is total number of treatment.

Here

r= 8

t=4

N= Total observation

= r*t

=32

The mean of these treatment can be calculate as

= 109 / 8

= 13.625

= 104 / 8

= 13

= 48/8

= 6

= 183 / 8

= 22.875

Now obtaining the grand mean ( )

=444 / 32

=13.875

Now calculating Sum of Squares.

Total Sum of Squares (TSS)

=(14-13.875)2+ (19-13.875)2 +(12-13.875)2 +...( 23-13.875)2 +( 2713.875)2

= 158735

The sum of squares between groups (SSB)

= ( 13.625-13.875)2+( 13-13.875)2+( 6-13.875)2+( 22.875-13.875)2

= 1150.75

sum of squares within groups (SSW)

SSW = TSS - SSB

=  1587.50 - 1150.75

= 436.75

Obtaining Degree of Freedom

df( between)= t-1

=3

df(within) = N -t

= 28

Calculating Mean Square of Within groups (MSW)

= 446.75 / 28

= 15.59821

Calculating Mean Square of Between groups (MSB)

= 1150.75 / 3

= 383.5833

Now obtaining the test statistic( F- statistic)

= 383.5833 / 15.5927

= 24.59149

The critical value is given as

= F0.05,3,28

From table

F0.05,3,28= 2.9467

Decision Rule:

Reject H0 null hypothesis if F (calculated) > F(critical value)

Here F (calculated)= 24.59149 > F(critical value) = 2.9467

Reject the null hypothesis.

Hence the mean weight loss due t. four diet plans: Life of Herbs (LOH), E-Mail-A-Meal (EAM), Weight-Lookers (WLK), and Low-Carbs (LOC) is significantly differenent

orchestra answered 1 year ago
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