4. a Researcher Has Constructed 80% Confidence Interval of Μ = 45 8, Using a Sample Of

4. A researcher has constructed 80% confidence interval of µ = 45 ± 8, using a sample of n = 25 scores.

a. What would happen to the width of the interval if the researcher had used a larger sample size?

b. What would happen to the width of the interval if the researcher had used 90% confidence instead of 80%?

c. What would happen to the width of the interval if the sample variance increased? (Assume other factors constant.)

(a) As the sample size increases, estimated standard error decreases so the width of interval will decrease.
(b) The higher the level of confidence the greater the width of the interval.
(c) As the variance increases, estimated standard error increases so the width of the interval will increases.

8. Problems 16 in chapter 9 described a study by Harlow (1959) in which infant monkeys were placed in cages with two artificial mothers. “One mother was made of wire mesh and had a bottle from which the infant could feed and the second mother of soft terry cloth but did not provide any food. Data for a sample of n = 9 monkeys showed that the infants spent an average an average of M = 15.3 hours per day with SS =216 with the terry cloth mother. Use the data to estimate how many hours per day would be spent with the terry cloth mother for the entire population of the infant monkeys. Make a point estimate and 80% confidence interval of the population mean.

(a) The point estimate is 15.3 hours
(b) s = √(SS/(n - 1)) = √(216/(9 - 1)) = 5.1962
SE = s/√n = 5.1962/√9 = 1.7321
t- score for 80% confidence with 8 degrees of freedom is 1.3968
ME = t * SE = 1.3968 * 1.7321 = 2.4193
The 80% CI is [x-bar - ME, x-bar + ME] = [15.3 - 2.42, 15.3 + 2.42] = [12.88 hours, 17.72 hours]

14. There is some evidence suggesting that you likely to improve your test score if you rethink and change answers on a multiple –choice exam (Johnston, 1975). To examine this phenomenon, a teacher gave the same final examto two sections of a psychology course. The students in one section were told to turn in their exams immediately after finishing, without changing any of the answers. In the other section, students were encouraged to reconsider each question and to change answers whenever they felt it was appropriate. The average score for the n = 20 students in the no- change section was M = 74.2 with SS = 460.6. The average for the n= 20 students in change section was M = 78.6 with SS = 512.2

a. Make a point estimate of the population mean difference mean difference between the test conditions.

b. Make a 95% confidence interval estimate of the mean difference between the two conditions.

c. Based on your answers to part b, would a two- tailed hypothesis test with α = .05 conclude that there is a significant difference between to conditions? (Is µ1-µ2= 0 an acceptable hypothesis?)

(a) M1 - M2 = MD
74.2 - 78.6 = -4.40 is the Point Estimate
(b) s1^2 = SS1/(n1 - 1) = 460.6/(20 - 1) = 24.24 and s2^2 = SS2/(n2 - 1) = 512.2/(20 - 1) = 26.96
Critical t- score for α = 0.05 and n1 + n2 - 2 = 20 + 20 - 2 = 38 degrees of freedom is 2.0244
Pooled SD, s = [{(n1 - 1) s1^2 + (n2 - 1) s2^2} / (n1 + n2 - 2)]
= [{(20 - 1) 24.24 + (20 - 1) 26.96} / (20 + 20 - 2)] = 5.0596
SE = s * {(1 /n1) + (1 /n2)} = 5.0596 * √{(1/20) + (1/20)} = 1.6
Margin of error, ME = t * SE = 2.0244 * 1.6 = 3.2390
LCL = (M1 - M2) - ME = -4.40 - 3.2390 = -7.64 and UCL = (M1 - M2) + ME = -4.40 + 3.2390 = -1.16
The 95% CI for the difference between means is [-7.64, -1.16]
(c) μ1 - μ2 = 0 lies outside the above confidence interval. Therefore, it appears that there is a significant difference between the two conditions.

12. The following data were obtained in a study using three separate samples to compare three different treatments.

Treatments

______

I II III

______

4 3 8 N = 12

3 1 4 G = 48

5 3 6 ∑Χ2 = 238

4 1 6

______

M = 4 M = 2 M = 6

T = 16 T = 8 T = 24

SS 2 SS = 4 SS = 8

a. Use an analysis of variance with α = .05 to determine whether there are significant difference among the treatments.

b. Compute the value for η2 for these data.

14. A researcher reports an F-ratio with df = 2, 24 for an independent –measures researcher study.

a. How many treatments conditions were compared in the study?

b. How many subjects participated in the entire study?

(a) 2 + 1 = 3 treatment conditions were compared
(b) 2 + 24 + 1 = 27 subjects participated in the study.

22. A developmental psychologist is examining the development of language skills from age 2 to age 5. Four different groups of children are obtained, one for each age with n = 15 children in each group. Each child is given a language skills assessments test. The resulting data were analyzed with an ANOVA to test to test for mean differences between age groups. The results of the ANOVA are presented in the following table. Fill in all missing values.

Source SS df MS F
Between Treatments 81 5 - 1 = 4 81/4 = 20.25 20.25/3.0545 =6.6296
Within Treatments 9 - 81 = 168 59 - 4 = 55 168/55 = 3.0545
Total 249 4 x 15 - 1 = 59

26. One possible explanation for why some birds migrate and others maintain year round residency in a singe location is intelligence. Specifically, birds with small brains, relative to their body size, are simply not smart enough to find food during the winter and must migrate to warmer climates where food is easily available (Sol, Lefebvre, & Rodriquez- Teijeiro, 2005). Birds with bigger brains, on the other hand, are more creative and can find food even when the weather turns harsh. Following are hypothetical data similar to actual research results. The numbers represent relative brain size for the individual birds in each sample.

______

Non- Short Long
Migrating Distance Distance
Migrants Migrant

______

18 6 4 N = 18

13 11 9 G = 180

19 7 5 ∑Χ2= 2150

12 9 6

16 8 5

12 13 7

______

M = 15 M = 9 M = 6

T = 90 T = 54 T = 36

SS = 48 SS = 34 SS = 16

______

a. Use AVOVA with α =.05 to determine whether there are any significant mean differences among three groups of birds.

b. Compute η2, the percentageof varianceexplained by the group difference, for these data.

c.Use the turkey HSD posttest to determine which groups are significant different.

(A) ANOVA:
(a) Ho: There is no difference between the treatment means; that is, 1 = 2 = 3
Ha: At least one among 1, 2 and 3 is different from the other two.
(b) One-way ANOVA with α = 0.05
Decision Rule: Reject H0 if the p- value < 0.05
(c) MegaStat output
One factor ANOVA
Mean / n / Std. Dev
15.0 / 6 / 3.10 / Non migrating
9.0 / 6 / 2.61 / Short distance migrants
6.0 / 6 / 1.79 / Long distance migrants
10.0 / 18 / 4.54 / Total
ANOVA table
Source / SS / df / MS / F / p-value
Treatment / 252.00 / 2 / 126.000 / 19.29 / .0001
Error / 98.00 / 15 / 6.533
Total / 350.00 / 17
(d) The p- value is 0.0001. Since this is < 0.05 we reject Ho and accept Ha.
(e) Conclusion: We conclude that at least one of the three means is different from the other two. That is, there are significant mean differences among the three groups of birds
(B) ^2 = Percentage of variance explained by the group differences
= (SS between groups / Total SS) * 100 = (252/350) * 100 = 72%
(C)
p-values for pair-wise t-tests
Long distance migrants / Short distance migrants / Non migrating
6.0 / 9.0 / 15.0
Long distance migrants / 6.0
Short distance migrants / 9.0 / .0602
Non migrating / 15.0 / 2.04E-05 / .0010
From Tukey's pair-wise p- values table, we see that the p- values are < 0.05 for the pairs Non-migrators vs Short distance migrators and Non-migrators vs Long distance migrators only. This means it is the Non-migrators category that is significantly different from the other two.