Question

1) You are interested in testing if there is a difference in weight-loss between 3 popular diet types; low calorie, low fat and low carb. Participants are randomly assigned one of the three groups or a control group. The control group will be told that are participating in a weight-loss study, but will not be following any particular diet. This is to study the placebo effect, weight-loss from just participating in the study.
At the start of the study,each participant's weight is recorded in pounds. After 8 weeks they are weighed again, and the weight difference is recorded. A positive value represents a weight loss while a negative value represents a weight gain. Test at 5% significance.

Low Calorie	Low Fat	Low Carb	Control
10.2	6.2	2.9	2.7
7.3	6	2.8	3.4
4.7	3.4	4.4	0.6
9	2	5.6	-1.4
2.7	3	3	0.7

What is the factor variable? Select an answer Diet Type Pounds Lost
What is the response variable? Select an answer Pounds Lost Diet Type
Test Statistic:
P-Value:
Decision Rule: Select an answer Accept the Null Reject the Null Fail to Reject the Null
Did Something Significant Happen? Select an answer Nothing Significant Happened Significance Happened
There Select an answer is is not  to conclude Select an answer The true average weight loss is the same for all diet types. At least one true average weight loss is different between the diet types.

2) Is a statistics class' delivery type a factor in how well students do on the final exam? The table below shows the average percent on final exams from several randomly selected classes that used the different delivery types.

Hybrid	Online	Face-to-Face
70	58	80
85	63	77
62	72	87
82	80	100
66	92	89
78	87	94
71	64	79
57

Assume that all distributions are normal, the three population standard deviations are all the same, and the data was collected independently and randomly. Use a level of significance of α=0.1α=0.1.

1. For this study, we should use Select an answer χ²GOF-Test 2-PropZTest TInterval 2-PropZInt 2-SampTTest 1-PropZTest 1-PropZInt T-Test 2-SampTInt χ²-Test ANOVA
2. 
   
3. Your friend Esmeralda helped you with the null and alternative hypotheses...
   H0: μ1=μ2=μ3H0: μ1=μ2=μ3
   H1:H1: At least one of the mean is different from the others.
   
4. The test-statistic for this data = (Please show your answer to 3 decimal places.)
5. 
   
6. The p-value for this sample =  (Please show your answer to 4 decimal places.)
7. 
   
8. The p-value is Select an answer greater than alpha less than (or equal to) alpha  αα
9. 
   
10. Base on this, we should Select an answer reject the null hypothesis fail to reject the null hypothesis accept the null hypothesis  hypothesis
11. 
    
12. As such, the final conclusion is that...
    • There is sufficient evidence to support the claim that course delivery type is a factor in final exam score.
    • There is insufficient evidence to support the claim that course delivery type is a factor in final exam score.

Answer 1

(1)

(a)

Question:

What is the factor variable?

Correct option:

Diet Type

(b)

Question:

What is the response variable?

Correct option:

Pounds Lost

(c)

From the given data,the following Table is calculated:

	Low calorie	Low fat	Low carb	Control	Total
N	5	5	5	5	20
	33.9	20.6	18.7	6.0	79.2
Mean	6.78	4.12	3.74	1.2	3.96
	267.71	99.0	75.97	21.66	464.34
Std. Dev.	3.0768	1.8794	1.228	1.9013	2.8164

From the above Table, ANOVA Table is calculated as follows:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F
Between treatments	78.22	3	78.22/3 = 26.0733	26.0733/ 4.5305 = 5.7551
Within treatments	72.488	16	72.488/16=4.5305
Total	150.708	19

Test Statisticis given by:

F = 26.0733/ 4.5305 = 5.7551

So,

Test Statistic = 5.7551

(d)

By Technology:

P - value = 0.0072

(e)

Correct option:

Reject the Null

(f)

Correct option:

Significance Happened

(g)

Correct option:

At least one true average weight loss is different between the diet types.

(2)

(a)

Correct option:

ANOVA

(b)

Correct option:

H0: μ1=μ2=μ3

:H1: At least one of the mean is different from the others.

(c)

From the given data,the following Table is calculated:

	Hybrid	Online	Face to face	Total
N	8	7	7	22
	571	516	606	1693
Mean	71.375	73.7143	86.5714	76.955
	41423	39046	52896	133365
Std. Dev.	9.7678	12.9707	8.5021	12.1125

From the above Table, ANOVA Table is calculated as follows:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F
Between treatments	969.9367	2	969.9367/2=484.9683	484.9683/ 111.1062 = 4.365
Within treatments	2111.0179	19	2111.0179/ 19 = 111.1062
Total	3080.9545	21

Test Statisticis given by:

F = 484.9683/ 111.1062 = 4.365

So,

Test Statistic = 4.365

(d)

By Technology:

P - value = 0.0276

(e)

Correct option:

Reject the Null hypothesis

(f)

Correct option:

There is sufficient evidence to support the claim that course delivery type is a factor in final exam score.