o Ch. 2, Practice Problems: 11, 12, 13, 16, and 21 o Ch. 3, Practice Problems: 14, 15, 22, and 25

11 - For the following scores, find the (a) mean, (b) median, (c) sum of squared deviations, (d) variance, and (e) standard deviation: 2, 2, 0, 5, 1, 4, 1, 3, 0, 0, 1, 4, 4, 0, 1, 4, 3, 4, 2, 1, 0

A. 2, 2, 0, 5, 1, 4, 1, 3, 0, 0, 1, 4, 4, 0, 1, 4, 3, 4, 2, 1, 0

Solution:

(a) Mean = sum/21 = 42/21 = 2

(b) Arrange the numbers in ascending order

0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 4, 4, 4, 4, 5

Median = middle number = 2

(c) Sum of squared deviations:

x x-mean(x-mean)^2

2 0 02 0 00 -2 45 3 91 -1 14 2 41 -1 13 1 10 -2 40 -2 41 -1 14 2 44 2 40 -2 41 -1 14 2 43 1 14 2 42 0 01 -1 10 -2 4

sum 56

Sum of squared deviation = 56

(d)Variance = Sum of squared deviation/n – 1 = 56/(21 – 1) = 56/20 = 2.8

(e)

Standard deviation = √variance = √2.8 = 1.673

12 - For the following scores, find the (a) mean, (b) median, (c) sum of squared deviations, (d) variance, and (e) standard deviation: 1,112; 1,245; 1,361; 1,372; 1,472

Solution:

(a) Mean = sum/21 = 6562/5 = 1312.4

(b) Arrange the numbers in ascending order

1112, 1,245, 1361, 1372, 1,472

Median = middle number = 1361

(c) Sum of squared deviations:

x x-mean (x-mean)^21112 -200.4 40160.161245 -67.4 4542.761361 48.6 2361.961372 59.6 3552.161472 159.6 25472.16

Sum 76089.2

Sum of squared deviations = 76089.2

(d)Variance = Sum of squared deviation/n – 1 = 76089.2/(5 – 1) = 76089.2/4 = 19022.3

(e)

Standard deviation = √variance = √19022.3 = 137.9

13 - For the following scores, find the (a) mean, (b) median, (c) sum of squared deviations, (d) variance, and (e) standard deviation: 3.0, 3.4, 2.6, 3.3, 3.5, 3.2

Solution:

(a) Mean = sum/6 = 19/6 = 3.17

(b) Arrange the numbers in ascending order

2.6, 3.0, 3.2, 3.3, 3.4, 3.5

Median = (3.2 + 3.3)/2 = 6.5/2 = 3.25

(c) Sum of squared deviations:

x x-mean (x-mean)^23 -0.17 0.0289

3.4 0.23 0.05292.6 -0.57 0.32493.3 0.13 0.01693.5 0.33 0.10893.2 0.03 0.0009

sum 0.5334

Sum of squared deviations = 0.5334

(d)Variance = Sum of squared deviation/n – 1 = 0.5334/(6 – 1) = 0.5334/5 = 0.1067

(e)

Standard deviation = √variance = √0.1067 = 0.3266

16 – A psychologist interested in political behavior measured the square footage of the desks in the official office of four U.S governors and of four chief executive officers (CEOs) of major U.S. corporations. The figures for the governors were 44, 36, 52, and 40 square feet. The figures for the CEOs were 32, 60, 48, 36 square feet. (a) Figure the mean and standard deviation for the governors and for the CEOs. (b) Explain what you have done to a person who has never had a course in statistics. (c) Note that ways in which the means and standard deviations differ, and speculate on the possible meaning of these differences, presuming that they are representative of U.S. governors and large corporations’ CEOs in general.

Solution:

a. For Governors:

Mean =  43Standard deviation =  6.831

For CEOs

Mean = (32+60+48+36)/4   = 44 Standard deviation = sqrt((32-44)^2 + (60 -44)^2 +(48 – 44)^2 +(36 – 44)^2)/3)= 12.649

b.Mean value can be found out by dividing the sum of all values within a group by total number of values within that group.

Standard deviation can be found by using the formula given below:

Where n is the number of data values in sample, X-bar is the mean and Xi are the data values.

The sample standard deviation (usually represented by S) measures the variability of data in a sample.c. The means are approximately the same for both which means that the basic average size (mean) does not vary much for a governor or a CEO. But there is a big difference in the standard deviations. It may occurred as CEO's have more freedom in choosing their desks while governors have some restrictions.

14 - On a standard measure of hearing ability, the mean is 300 and the standard deviation is 20. Give the Z scores for persons who score (a) 340, (b) 310, and (c) 260. Give the raw scores for persons whose Z scores on this test are (d) 2.4, (e) 1.5, (f) 0, and (g) –4.5.

Solution:

We will use the formula:

Given = 300, = 20

a. Here X = 340

Therefore, Z = = 2

b. Here X = 310

Therefore, Z = = 0.5

c. Here X = 260

Therefore, Z = = -2

For parts (d) to (g) use the formula

X = + Z

d. Here Z = 2.4

Therefore, X = 300 + 20 x 2.4 = 348

e. Here Z = 1.5

Therefore, X= 300 + 20 x 1.5 = 330

f. Here Z = 0

Therefore, X= 300 + 20 x 0 = 300

g. Here Z = -4.5

Therefore, X= 300 + 20 x (-4.5) = 210

15 - A person scores 81 on a test of verbal ability and 6.4 on a test of quantitative ability. For the verbal ability test, the mean for people in general is 50 and the standard deviation is 20. For the quantitative ability test, the mean for people in general is 0 and the standard

deviation is 5. Which is this person’s stronger ability: verbal or quantitative? Explain your answer to a person who has never had a course in statistics.

Solution:

We will find the z-sores for both the abilities. The formula for z-score = (score - mean)/standard deviation

Find the z-score for verbal ability = (score - mean)/standard deviation = (81 - 50)/20 = 31/20 = 1.55

Find the z-score for quantitative ability = (score - mean)/standard deviation = (6.4 - 0)/5 = 6.4/5 = 1.28

The z-score of verbal ability is greater than z-score of quantitative ability.

Thus, person's stronger ability is verbal ability.

Since the z-score of verbal ability is greater than z-score of quantitative ability therefore person's stronger ability is verbal ability.

22 - Suppose you want to conduct a survey of the attitude of psychology graduate students studying clinical psychology towards psychoanalytic methods of psychotherapy. One approach would be to contact every psychology graduate student you know and ask them to fill out a questionnaire about it. (a) What kind of sampling method is this? (b) What is a major limitation of this kind of approach?

Solution:

(a) This type of sampling will be convenience sampling. A convenience sample chooses the individuals that are easiest to reach or sampling that is done easy. In this case the psychology graduate student sample is easiest sample to get.

(b) The major limitation of this sampling is that it is biased. Since convenience sampling does not represent the entire population so it is considered bias. Analysis carried out on the basis of such sampling may be incorrect or inaccurate.

25 - You are conducting a survey at a college with 800 students, 50 faculty members, and 150 administrative staff members. Each of these 1,000 individuals has a single listing in the campus phone directory. Suppose you were to cut up the directory and pull out one listing at random to contact. What is the probability it would be (a) a student, (b) a faculty member, (c) an administrative staff member, (d) a faculty or administrative staff member, and (e) anyone except an administrative staff member? (f) Explain your answers to someone who has never had a course in statistics.

Solution:(a)P(a student) = 800/1000 =0.8

(b) P(a faculty member) = 50/1000 = 0.05

(c) P(an administrative staff member) = 150/1000 = 0.15

(d)P(a faculty or an administrative staff member) = (50 + 150)/1000 = 200/1000 = 0.2

(e)P(anyone except an administrative staff member) = 850/1000 = 0.85

Ch 4 13,16,20Ch 7 14Ch 8 18Ch 9 17Ch 11 11 & 12

13 - List the five steps of hypothesis testing and explain the procedure and logic of each.

Answer:

First step:

The first step in hypothesis testing is to specify the null hypothesis (H0) and the alternative hypothesis (H1).

Second step:

In this step we have to select significance level. If significance level is not given then 0.05 or the 0.01 level is used.

Third step:

Calculate the test statistic. The third step is to calculate a statistic analogous to the parameter specified by the null hypothesis.

Fourth step:

The fourth step is to calculate the p-value which is the probability of the test statistic and find the rejection region.

Fifth step:

The fifth and final step is to draw and describe conclusions. The probability value computed in 4th step is compared with the significance level selected in step 2. If the probability is less than or equal to the significance level, then the null hypothesis is rejected. If the probability is greater than the significance level then the null hypothesis is not rejected.

16 - Based on the information given for each of the following studies, decide whether to reject the null hypothesis. For each, give the Z-score cutoff (or cutoffs) on the comparison distribution at which the null hypothesis should be rejected, the Z score on the comparison distribution for the sample score, and your conclusion. Assume that all populations are normally distributed.

PopulationSample ScorepTails of Testµ s

A 5 1 7 .05 1 (high predicted)B 5 1 7 .05 2C 5 1 7 .01 1 (high predicted)D 5 1 7 .01 2

Solution:Study A:z –score = (7 – 5)/1 = 2At .05 significance level, the z-score is + 1.64z-score cut-off value = +1.64Since 2 > 1.64 so reject Null hypothesis.Study B:z –score = (7 – 5)/1 = 2At .05 significance level for two-tailed test, the z-score is ±1.96.z-score cut-off value = ±1.96

Since 2 > +1.96 so reject Null hypothesis.Study C:z –score = (7 – 5)/1 = 2At .01 significance level, the z-score is +2.33z-score cut-off value = +2.33Since 2 < 2.33 so fail to reject Null hypothesis.Study D:z –score = (7 – 5)/1 = 2At .01 significance level for two-tailed test, the z-score is ±2.58.

z-score cut-off value = ±2.58Since -2.58<2<2.58 so fail to reject Null hypothesis.

20 - A researcher predicts that listening to music while solving math problems will make a particular brain area more active. To test this, a research participant has her brain scanned while listening to music and solving math problems, and the brain area of interest has a percent signal change of 58. From many previous studies with this same math-problems procedure (but not listening to music), it is known that the signal change in this brain area is normally distributed with a mean of 35 and a standard deviation of 10. Using the .01 level, what should the researcher conclude? Solve this problem explicitly using all five steps of hypothesis testing, and illustrate your answer with a sketch showing the comparison distribution, the cutoff (or cutoffs), and the score of the sample on this distribution. Then explain your answer to someone who has never had a course in statistics (but who is familiar with mean, standard deviation, and Z scores).

Solution:

Null Hypothesis: H0: μ = 35

Alternate Hypothesis: H1: μ ≠ 35

Significance level: α = 0.01

Critical value = ±2.58

Test statistic:

Z = (xbar-μ)/(σ) = (58-35)/(10) = 2.3

Conclusion:

Since 2.3 < 2.58 we fail to reject H0. Researcher should not predict.

Here is a sketch of the critical region:

The above diagram shows the rejection region. We can see that the z-statistic 2.3 does not lie in the rejection region so we fail to reject Null Hypothesis and conclude that the average percent signal change is 35.

14 - Evolutionary theories often emphasize that humans have adapted to their physical environment. One such theory hypothesizes that people should spontaneously follow a 24- hour cycle of sleeping and waking—even if they are not exposed to the usual pattern of sunlight. To test this notion, eight paid volunteers were placed (individually) in a room in which there was no light from the outside and no clocks or other indications of time. They could turn the lights on and off as they wished. After a month in the room, each individual tended to develop a steady cycle. Their cycles at the end of the study were as follows: 25, 27, 25, 23, 24, 25, 26, and 25.

Using the 5% level of significance, what should we conclude about the theory that 24 hours is the natural cycle? (That is, does the average cycle length under these conditions differ significantly from 24 hours?) (a) Use the steps of hypothesis testing. (b) Sketch the distributions involved. (c) Explain your answer to someone who has never taken a course in statistics.

Solution:Solution:

(a)Null hypothesisH0: = 24 hours

Alternative hypothesis:H1: ≠ 24 hours

Df = 7Critical t value = ±2.36

Sample Mean = 25

Standard deviation = 1.195

The test statistic used is

p-value = 0.049867231

Conclusion: since calculated p-value 0.049867231 is slightly less than 0.05(significance level) therefore we reject the null hypothesis.

Thus there is enough evidence to support the claim that the average cycle length under experimental conditions is significantly different from 24 hours.

(b)Sketch is shown below:

18 - Twenty students randomly assigned to an experimental group receive an instructional program; 30 in a control group do not. After 6 months, both groups are tested on their knowledge. The experimental group has a mean of 38 on the test (with an

estimated population standard deviation of 3); the control group has a mean of 35 (with an estimated population standard deviation of 5). Using the .05 level, what should the experimenter conclude? (a) Use the steps of hypothesis testing, (b) sketch the distributions involved, and (c) explain your answer to someone who is familiar with the t test for a single sample, but not with the t test for independent means.

Solution:

AnswerThe null hypothesis tested isH0: There is no significant difference in the mean score of the two groups H1: There is significant difference in mean knowledge of two groups.

The test statistic t = 2.648204Critical value = 2.01

Decision rule: Reject the null hypothesis, if the calculated value of t is greater than the critical value of t.

Conclusion: Reject the null hypothesis. There is enough evidence to support that there is significant difference in mean knowledge of two groups.

We have used t test to examine whether there is significant difference in mean knowledge of two groups. The test concludes that there is significant difference in mean knowledge of two groups.

17 - Do students at various colleges differ in how sociable they are? Twenty-five students were randomly selected from each of three colleges in a particular region and were asked to report on the amount of time they spent socializing each day with other students. The results for College X was a mean of 5 hours and an estimated population variance of 2 hours; for College Y, M = 4, S2 = 1.5; and for College Z, M = 6, S2 = 2.5. What should you conclude? Use the .05 level. (a) Use the steps of hypothesis testing, (b) figure the effect size for the study; and (c) explain your answers to (a) and (b) to someone who has never had a course in statistics.

Answer

Null hypothesis:H0: There is no significant difference in the mean amount of time spent for socializing each day with other students by student from 3 colleges.

Alternative hypothesis:H1: There is significant difference in the mean amount of time spent for socializing each day with other students by student from 3 colleges.

The test statistic used is F test (ANOVA). Decision rule: Reject the null hypothesis, if the value of test statistic is greater than the critical value of F.Details (Please see the excel file for calculations)

SS df MSS F F criticalT SS 50 2 25 12 3.123907ESS 150 72 2.083333Total SS 200 74 2.702703

Conclusion: Reject the null hypothesis. The sample provides enough evidence to support the claim that there is significant difference in the mean amount of time spent for socializing each day with other students by student from 3 colleges.

Effect Size

Eta squared is the proportion of the total variance that is attributed to an effect. It is calculated as the ratio of the effect variance (SSeffect) to the total variance (SStotal)

2 = SSeffect / SStotal =50/200 =0.25

ANOVA test is used test whether there is difference in the mean amount of time spent for socializing each day with other students by student from 3 colleges. The Value of test

statistic is greater than the critical value indicate that there is significant difference in the mean values.

The effect size for the study is 0.25. Thus 25% of the total variance in the amount of time spent for socializing each day with other students is attributed to the college.

11 - Make up a scatter diagram with 10 dots for each of the following situations:

(a) Perfect positive linear correlation

Solution:

2 43 65 101 24 86 127 148 169 18

10 20

r = +1

0

5

10

15

20

25

0 2 4 6 8 10 12

Series1

(b) large but not perfect positive linear correlation

Solution:

2 43 5

5 81 14 76 127 94 88 151 2

r = +0.95108

02468

10121416

0 2 4 6 8 10

Series1

(c) small positive linear correlation

2 21 52 47 54 80 61 73 151 21 8

r = + 0.138596

02468

10121416

0 2 4 6 8

(d) large but not perfect negative linear correlation

Solution:

2 63 65 81 84 36 17 1

10 02 74 4

r = -0.81927

0123456789

0 2 4 6 8 10 12

Series1

(e) no correlation

Solution:

10 6020 5015 3012 1230 3014 102 83 204 8

20 2

0

10

20

30

40

50

60

70

0 10 20 30 40

Series1

(f) clear curvilinear correlation

Solution:

1 12 43 94 165 25

12 14415 22520 4009 81

10 100

050

100150200250300350400450

0 5 10 15 20 25

Series1

For problems 12, do the following: (a) Make a scatter diagram of the scores; (b) describe in words the general pattern of correlation, if any; (c) figure the correlation coefficient; (d) figure whether the correlation is statistically significant (use the .05 significance level, two-tailed); (e) explain the logic of what you have done, writing as if you are speaking to someone who has never heard of correlation (but who does understand the mean, deviation scores, and hypothesis testing); and (f) give three logically possible directions of causality, indicating for each direction whether it is a reasonable explanation for the correlation in light of the variables involved (and why).

12 - Four research participants take a test of manual dexterity (high scores mean better dexterity) and an anxiety test (high scores mean more anxiety). The scores are as follows.

Person Dexterity Anxiety1 1 102 1 83 2 44 4 –2

Solution:

(a)

SCATTER DIAGRAM

0

2

4

6

8

10

12

0 1 2 3 4 5

DEXTERITY

AN

XIET

Y

(b)There is negative correlation between dexterity and anxiety

(c)

Correlation coefficient r = -0.9037

(d)

Ho: Correlation coefficient is greater than or equal to zeroHa: Correlation coefficient is less than zero

Test statistic t = -2.98481p-value = 0.048152

Since p-value is less than 5% we reject Ho and conclude that correlation coefficient is negative.

(e)If there is increase in dexterity there is decrease in anxiety.