1.) Write the null and alternative hypotheses for the following items 10, 12, 16, 18, and 20.

10.

12.

16.

18.

20.

Pease use the information found below this line

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10.) Research has shown that losing even on nights sleep can have a significant effect on performance of complex tasks such as problem solving (Linde & Bergstroem, 1992). To demonstrate this phenomenon, a sample of college students was given a problem-solving task at noon on one day and again at noon on the following day. The students were not permitted any sleep between the two tests. For each student, the difference between the first and second score was recorded. For this sample, the students averaged MD.7 points better on the first test with a variance of s^ for the difference scores.

A) Do the data indicated a significant change in problem solving ability? Use a two-tailed test with =.05

Answer:

Ho: d = 0 vs Ha: d 0

Df = n - 1 = 25 - 1 = 24

SE of the differences = s/n = 8/25 = 1.6

t = (Md - d)/SE = (4.7 - 0)/1.6 = 2.9375

Critical t- score for = 0.05 and Df = 24 is 2.0639

Since 2.9375 > 2.0639, we reject Ho and accept Ha

Conclusion: It appears that there is a significant change in the problem-solving ability.

B) Compute an estimated cohens d to measure the size of the effect.

Answer:

Cohens d = Md/s = 4.7/8 = 0.5875

r^2 = t^2 / (t^2 + df) = 2.9375^2 / (2.9375^2 + 24) = 0.2645 (26.45%)

12.) How would you react to doing much worse on an exam than you expected? There is some evidence to suggest that most individuals believe that they can cope with this kind of problem better than their fellow students (Igou, 2008). In the study, participants read a scenario of a negative event and were asked to use a 10-point scale to rate how it would affect their immediate well-being (-5 strongly worsen to +5 strongly improve). Then they were asked to imagine the event from the perspective of an ordinary fellow student and rate how it would affect that person. The difference between the two ratings was recorded. Suppose that a sample of participants produced a mean difference of MD.28 points (self rated higher) with a standard deviation of .50 for the difference scores.

A) Is this result sufficient to conclude that there is a significant difference in the ratings for self versus others? Use a two-tailed test with =.05

Answer:

Hypotheses are:

Sample standard deviation for the difference of the data is

Standard error for this data is

So value of t is

Degree of freedom of the test is

Since = .05, so critical values for are -2.0639 and 2.0639.

Since test statistics value is greater than 2.0639 so we reject the null hypothesis.

B) Compute r^2 and estimate Cohens d to measure the size of the treatment effect.

Answer:

Cohen's d:

C) Write a sentence describing the outcome of the hypothesis test and the measure of effect size as it would appear in a research report.

Answer:

With .267 and p-value < 0.05, I discard the null hypothesis and conclude that difference is significant. The current effect size is large.

16.) A researcher for a cereal company wanted to demonstrate the health benefits of eating oatmeal. A sample of 9 volunteers was obtained and each participants ate a fixed diet without any oatmeal for 30 days. At the end of the 30-day period, cholesterol was measured for each individual. Then the participants began a second 30-day period in which they repeated exactly the same diet except that they added 2 cups of oatmeal each day. After the second 20-day period, cholesterol levels were measured again and the researcher recorded the difference between the two scores for each participant. For this sample, cholesterol scores averaged points lower with the oatmeal diet with for the difference scores.

A) Is the data sufficient to indicate a significant change in cholesterol level? Use a two-tailed test with =.01

Answer:

Ho: no significant difference in the cholesterol level. uD = 0

H1: significant difference in the cholesterol level. uD =/= 0

Alpha = 0.01

T(a/2,n-1) = +- 3.355

Test statistic, to = (Md-uD)/(sMD)

s^2 = ss/(n-1) = 538/8 = 67.25

s Md = sqrt(s^2/n) = sqrt(67.25/9) = 2.7335

hence, to = (16-0)/2.7335 = 5.853

since, to > t(a/2,n-1), I discard ho at 5% level of significance and conclude that there is significant difference in the cholesterol level. uD =/= 0

B) Compute r^2, the percentage of variance accounted for by the treatment, to measure the size of the treatment effect.

Answer:

r^2 = t^2/(t^2+df) = 5.853^2/(5.853^2+9) = 79.19%

C) Write a sentence describing the outcome the hypothesis test and the measure of effect size as it would appear in a research report.

Answer:

With .853 and p-value < 0.05, I reject ho at 5% level of significance and conclude that there is significant difference in the cholesterol level. uD =/= 0. The 79.19% of the variance accounted for by the treatment. To measure the size of the treatment effect.

18.) One of the primary advantages of a repeated-measures design, compared to independent-measures, is that it reduces the overall variability by removing variance caused by individual differences. The following data are from a research study comparing two treatment conditions.

For T1, , ss1 = 34, s1^/.8

For T2,, , s2^/

Diff, , , sd^/

A) Assume that the data are from an independent measures study using two separate samples, each with participants. Compute the pooled variance and the estimated standard error for the mean difference.

Answer:

= (5*6.8 + 5*6)/(6+6-2) = 64/10 = 6.4

SE for diff in means = sp = sqrt(sp^2) = sqrt(6.4) = 2.53

B) Now assume that the data are from a repeated measures study using the same sample of participants in both treatment conditions. Compute the variance for the sample of difference scores and the estimated standard error for the mean difference. (You should find that the repeated measures design substantially reduces the variance and the standard error.)

Treatment 1 Treatment 2 Treatment 3

10 13 3

12 12 0

8 10 2

6 10 4

5 6 1

7 9 2

M = 8 M = 10 MD = 2

SS = 34 SS = 30 SS = 10

Answer:

sD^2 = SSd/(n-1)= 10/5

Sd = sqrt(sD^2 ) = sqrt(2) = 1.41

The repeated measures design substantially reduces the variance and SE.

20.) A researcher uses a matched-subjects design to investigate whether single people who own pets are generally happier than singles without pets. A mood inventory questionnaire is administered to a group of 20- to 29-year-old non-pet owners and a similar age group of pet owners. The pet owners are matched one to one with the non-pet owners for income, number of close friendships, and general health. The data are as follows:

Matched Pair Non-Pet Owner Pet Owner

A 12 14

B 8 7

C 10 13

D 9 9

E 7 13

F 10 12

A) Is there a significant difference in the mood scores for non-pet owners versus pet owners? Test with =.05 for two tails.

Answer:

Ho: no significant diference in the mean mood scores for non pet owners and pet owners. ud = 0

H1: significant diference in the mean mood scores for non pet owners and pet owners. ud =/= 0

Alpha = 0.05

Mean(d) = -5

Sd(d) .899

T = mean(d)/(sd(d)/sqrt(n))

= -5/(4.899/sqrt(6))

= -2.5

p-value = 0.0545

Since the p-value is greater than alpha (0.05), I fail to reject ho and conclude that no significant difference in the mean mood scores for non pet owners and pet owners. ud = 0.

B) Construct the 95% confidence interval to estimate the size of the mean difference in mood between the population of pet owners and the population of non-pet owners. (You should find that a mean difference of D is an acceptable value, which is consistent with the conclusion from the hypothesis test.)

Answer:

95% confidence interval = mean(d) t(a/2,n-1)*(sd(d)/sqrt(n))

Lower = -5-2.57*(4.899/sqrt(6)) = -10.14

Upper = -5+2.57*(4.899/sqrt(6)) = 0.14