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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 |0 |

|2 |0 |0 |

|0 |-2 |4 |

|5 |3 |9 |

|1 |-1 |1 |

|4 |2 |4 |

|1 |-1 |1 |

|3 |1 |1 |

|0 |-2 |4 |

|0 |-2 |4 |

|1 |-1 |1 |

|4 |2 |4 |

|4 |2 |4 |

|0 |-2 |4 |

|1 |-1 |1 |

|4 |2 |4 |

|3 |1 |1 |

|4 |2 |4 |

|2 |0 |0 |

|1 |-1 |1 |

|0 |-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)^2 |

|1112 |-200.4 |40160.16 |

|1245 |-67.4 |4542.76 |

|1361 |48.6 |2361.96 |

|1372 |59.6 |3552.16 |

|1472 |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)^2 |

|3 |-0.17 |0.0289 |

|3.4 |0.23 |0.0529 |

|2.6 |-0.57 |0.3249 |

|3.3 |0.13 |0.0169 |

|3.5 |0.33 |0.1089 |

|3.2 |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 =  43

Standard 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.

[pic]

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:

[pic]

Given [pic]= 300, [pic] = 20

a. Here X = 340

Therefore, Z = [pic] = 2

b. Here X = 310

Therefore, Z = [pic] = 0.5

c. Here X = 260

Therefore, Z = [pic] = -2

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

X = [pic]+ Z[pic]

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 13,16,20

Ch 7 14

Ch 8 18

Ch 9 17

Ch 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.

|Population |Sample Score |p|Tails of Test |

|µ |s | | | |

|A |5 |1 |7|.05 |1 (high predicted) |

|B |5 |1 |7|.05 |2 |

|C |5 |1 |7|.01 |1 (high predicted) |

|D |5 |1 |7|.01 |2 |

Solution:

Study A:

z –score = (7 – 5)/1 = 2

At .05 significance level, the z-score is + 1.64

z-score cut-off value = +1.64

Since 2 > 1.64 so reject Null hypothesis.

Study B:

z –score = (7 – 5)/1 = 2

At .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 = 2

At .01 significance level, the z-score is +2.33

z-score cut-off value = +2.33

Since 2 < 2.33 so fail to reject Null hypothesis.

Study D:

z –score = (7 – 5)/1 = 2

At .01 significance level for two-tailed test, the z-score is ±2.58.

z-score cut-off value = ±2.58

Since -2.58 ................
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