Self-efficacy is a general concept that measures how well we think we can contro
ID: 3337935 • 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:
x
(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 not Normal because the sample included only women. The distribution is Normal because the standard deviation is smaller than the mean.The distribution is Normal because the sample sizes are large.The distribution is Normal because the sample was randomly assigned.
(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 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 not be appropriate because we do not have Normally distributed data.The t procedures should be appropriate because we have Normally distributed data.
(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 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 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 decrease scores on the test. The one-sided alternative allows for the possibility that the intervention might have had a positive 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.)
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.
Group nx
s Intervention 163 4.17 1.19 Control 217 3.62 1.12Explanation / Answer
a.
The distribution is Normal because the standard deviation is smaller than the mean.The distribution is Normal because the sample sizes are large.The distribution is Normal because the sample was randomly assigned
Given that,
mean(x)=4.17
standard deviation , s.d1=1.19
number(n1)=163
y(mean)=3.62
standard deviation, s.d2 =1.12
number(n2)=217
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.654
since our test is right-tailed
reject Ho, if to > 1.654
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =4.17-3.62/sqrt((1.4161/163)+(1.2544/217))
to =4.5725
| to | =4.5725
critical value
the value of |t | with min (n1-1, n2-1) i.e 162 d.f is 1.654
we got |to| = 4.57249 & | t | = 1.654
make decision
hence value of | to | > | t | and here we reject Ho
p-value:right tail - Ha : ( p > 4.5725 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
c.
null, Ho: u1 = u2
alternate, H1: u1 > u2
b.
test statistic: 4.5725
critical value: 1.654
decision: reject Ho
p-value: 0
d.
Critical Value
The Value of t at 0.05 LOS with n-1 = 216 is -1.652
P-Value : Left Tail - Ha :( P < 4.5725) = 1
Critical Value
The Value of t at 0.05 LOS with n-1 = 216 is +1.652
P-Value : Right Tail - Ha :( P > 4.5725) = 0
Critical Value
The Value of t at 0.05 LOS with n-1 = 162 is -1.654
P-Value : Left Tail - Ha :( P < 4.5725) = 1
Critical Value
The Value of t at 0.05 LOS with n-1 = 162 is +1.654
P-Value : Right Tail - Ha :( P > 4.5725) = 0
e.
TRADITIONAL METHOD
given that,
mean(x)=4.17
standard deviation , s.d1=1.19
number(n1)=163
y(mean)=3.62
standard deviation, s.d2 =1.12
number(n2)=217
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((1.416/163)+(1.254/217))
= 0.12
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 162 d.f is 1.975
margin of error = 1.975 * 0.12
= 0.238
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (4.17-3.62) ± 0.238 ]
= [0.312 , 0.788]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=4.17
standard deviation , s.d1=1.19
sample size, n1=163
y(mean)=3.62
standard deviation, s.d2 =1.12
sample size,n2 =217
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 4.17-3.62) ± t a/2 * sqrt((1.416/163)+(1.254/217)]
= [ (0.55) ± t a/2 * 0.12]
= [0.312 , 0.788]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [0.312 , 0.788] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.