Self-efficacy is a general concept that measures how well we think we can contro
ID: 3201338 • Letter: S
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 166 4.12 1.19
Control 214 3.66 1.12
(a) Do you think that these data are Normally distributed? Explain why or why not.
The distribution is Normal because the sample sizes are large.
The distribution is Normal because the sample was randomly assigned.
The distribution is not Normal because the sample included only women.
The distribution is not Normal because all scores are integers.
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 be appropriate because we have Normally distributed data.
The t procedures should not be appropriate because the sample sizes are not large enough.
The t procedures should not be appropriate because we do not have Normally distributed data.
The t procedures should not be appropriate because the two groups are different sizes.
The t procedures should be appropriate because we have two large samples with no outliers.
(c) Describe appropriate null and alternative hypotheses.
H0: Intervention 2; Ha: Intervention < Control (or Intervention = Control)
H0: Intervention = 2; Ha: Intervention > Control (or Intervention = Control)
H0: Intervention = 2; Ha: Intervention > Control (or Intervention < Control)
H0: Intervention 2; Ha: Intervention > Control (or Intervention = Control)
H0: Intervention = 2; Ha: Intervention < Control (or Intervention 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 increase scores on the test. The two-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 positive 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 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 increase scores on the test. The one-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. (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.
(_______ , ________)
(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 will generalize well to all other areas of the country.
The results for this sample may not generalize well to other areas of the country.
Could you please walk ne through the process for solving the T value, P value and df. I get really confused with this process, Thank You.
Explanation / Answer
Solution:-
Group n x s
Intervention 166 4.12 1.19
Control 214 3.66 1.12
a) The distribution is Normal because the sample sizes are large.
b) The t procedures should not be appropriate because the two groups are different sizes.
c) H0: Intervention = 2; Ha: Intervention > Control (or Intervention < Control)
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.
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1< 2
Alternative hypothesis: 1 > 2
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
d) SE = sqrt[(s12/n1) + (s22/n2)]
SE = 0.11997
DF = 224.06 (By using Calculator)
D.F = 224
t = [ (x1 - x2) - d ] / SE
t = 3.834
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.
The observed difference in sample means (46) produced a t statistic of 3.834. We use the t Distribution Calculator to find P(t > 0.3843) = 0.000082
Therefore, the P-value in this analysis is 0.000082
Interpret results. Since the P-value (0.000082) is less than the significance level (0.05), we have to reject the null hypothesis.
We reject H0 and conclude that the intervention increased test scores.
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