Question

Self-efficacy is a general concept that measures how well we think we can control different situations. A multimedia program designed to improve dietary behavior among low-income women was evaluated by comparing women who were randomly assigned to intervention and control groups. Participants were asked, "How sure are you that you can eat foods low in fat over the next month?" The response was measured on a five-point scale with 1 corresponding to "not sure at all" and 5 corresponding to "very sure." Here is a summary of the self-efficacy scores obtained about 2 months after the intervention:

Group	n	x	s
Intervention	169	4.15	1.19
Control	217	3.66	1.12

(a) Do you think that these data are Normally distributed? Explain why or why not.

____The distribution is not Normal because all scores are integers.

____The distribution is Normal because the sample was randomly assigned.   

____The distribution is not Normal because the sample included only women.

____The distribution is Normal because the sample sizes are large.

____The distribution is Normal because the standard deviation is smaller than the mean.

(b) Is it appropriate to use the two-sample t procedures that we studied in this section to analyze these data? Give reasons for your answer.

___The t procedures should not be appropriate because the two groups are different sizes.

___The t procedures should not be appropriate because we do not have Normally distributed data.    

___ The t procedures should not be appropriate because the sample sizes are not large enough.

___The t procedures should be appropriate because we have two large samples with no outliers.

___The t procedures should be appropriate because we have Normally distributed data.

(c) Describe appropriate null and alternative hypotheses.

___H₀: μ_Intervention = μ_Control; H_a: μ_Intervention < μ_Control (or μ_Intervention ≠ μ_Control)

___H₀: μ_Intervention ≠ μ_Control; H_a: μ_Intervention < μ_Control (or μ_Intervention = μ_Control)

___ H₀: μ_Intervention = μ_Control; H_a: μ_Intervention > μ_Control (or μ_Intervention < μ_Control)

___H₀: μ_Intervention ≠ μ_Control; H_a: μ_Intervention > μ_Control (or μ_Intervention = μ_Control)

___H₀: μ_Intervention = μ_Control; H_a: μ_Intervention > μ_Control (or μ_Intervention ≠ μ_Control)

Some people would prefer a two-sided alternative in this situation while others would use a one-sided significance test. Give reasons for each point of view.

___The one-sided alternative reflects the researchers' (presumed) belief that the intervention would decrease scores on the test. The two-sided alternative allows for the possibility that the intervention might have had a positive effect.

___The two-sided alternative reflects the researchers' (presumed) belief that the intervention would increase scores on the test. The one-sided alternative allows for the possibility that the intervention might have had a negative effect.    

___The one-sided alternative reflects the researchers' (presumed) belief that the intervention would decrease scores on the test. The two-sided alternative allows for the possibility that the intervention might have had a negative effect.

___The two-sided alternative reflects the researchers' (presumed) belief that the intervention would decrease scores on the test. The one-sided alternative allows for the possibility that the intervention might have had a positive effect.

___The one-sided alternative reflects the researchers' (presumed) belief that the intervention would increase scores on the test. The two-sided alternative allows for the possibility that the intervention might have had a negative effect.

(d) Carry out the significance test using a one-sided alternative. Report the test statistic with the degrees of freedom and the P-value. (Use μ_Intervention − μ_Control. Round your test statistic to three decimal places, your degrees of freedom to the nearest whole number, and your P-value to four decimal places.)

t	=
df	=
P-value	=

Write a short summary of your conclusion.

___We do not reject H₀ and conclude that the intervention had no significant effect on test scores.

___We reject H₀ and conclude that the intervention increased test scores.    

(e) Find a 95% confidence interval for the difference between the two means. Compare the information given by the interval with the information given by the significance test.

( ____ , ____ ) Answers

  ,
(f) The women in this study were all residents of Durham, North Carolina. To what extent do you think the results can be generalized to other populations?

___The results for this sample may not generalize well to other areas of the country.

___The results for this sample will generalize well to all other areas of the country.    

THANK YOU

Answer 1

(a)

Correct option:

The distribution is Normal because the sample sizes are large.

(b)

Correct option:

The t procedures should be appropriate because we have two large samples with no outliers.

(c)

Correct option:

H₀: μ_Intervention = μ_Control; H_a: μ_Intervention > μ_Control (or μ_Intervention ≠ μ_Control)

Correct option:

The one-sided alternative reflects the researchers' (presumed) belief that the intervention would increase scores on the test. The two-sided alternative allows for the possibility that the intervention might have had a negative effect.

(d)

Test Statistic is given by:

t = (4.15 - 3.66)/0.1181

= 4.149

ndf = n1 + n2 - 2

= 169 + 217 - 2

= 384

By Technology, P - Value = 0.0000

So,

Answer is:

t =	4.149
df =	384
P - value	0.0000

Correct option:

We reject H₀ and conclude that the intervention increased test scores.    

(e)

From Table, critical values of t = 1.9662

Confidence Interval:

(4.15 - 3.66) (1.9662 X 0.1181)

= 0.49 0.2322

= ( 0.2578 ,0.7222)

So,

Answer is:

( 0.2578 ,0.7222)

Since all values in the Confidence Interval are positive it confirms the conclusion that the intervention increased test scores.    

(f)

Correct option:

The results for this sample may not generalize well to other areas of the country.