Chapter 4

Chapter4

  1. What does a measure of central tendency indicate?

Ameasure of central tendency indicates the location of a distributionon a variable. They include the mean, median and the mode

13.For the following data, compute (a) the mean and (b) the mode.

55, 57, 59, 58, 60, 57, 56, 58, 61, 58, 59.

  1. Mean =total sum/number of variable

=(55+57+59+58+60+57+56+58+61+58+59)/11

=638/11 = 58

b.Mode = the variable that appears the most. From the data, 55 appearsonly once, 57 appears twice, 59 appears twice, 58 appears thrice 60appears once 56 once, and 61 once. Therefore the most representeddata is 58. Hence mode = 58

15.A researcher collected the following sets of data. For each, indicatethe measure of central tendency she should compute:

(a). the following IQ scores: 60, 72, 63, 83, 68, 74, 90, 86, 74, and 80

Mean= (60+72+63+83+68+74+90+86+74+80)/ 10 = 750/10 = 75

Median= the middle variable when arranged in ascending order

60,63, 68, 72, 74, 74, 80, 83, 86, 90

Middlenumbers are 74, 74

Thereforemedian = (74 +74)/2 = 74

Mode= 74 since it appears twice

(b)the following error scores: 10, 15, 18, 15, 14, 13, 42, 15, 12, 14,and 42

Mean= (10 +15+ 18+15+14+13+42+15+12+14+42)/11

=210/11 = 19.09

Median= middle number, 10, 12, 13, 14, 14, 15, 15, 15, 18, 42, 42

Median= 15

Mode= 15 which appears 3 times

(c)the following blood types: A-,A-, O, A+, AB-, A+, O,O, O, and AB+

Mean= (2A- + 4 O + 2A+ + AB- + AB+)10

Mode= blood group O

Median= O O O O A- A- A+ A+ AB- AB+= blood group A-

(d)the following grades: B, D, C, A, B, F, C, B, C, D, and D.

Mean= (3 B + 3D + 3C + A + F)/11

Median:A B B B C C C D D D F = C

Mode= (B +C +D)/3

18.In a normal distribution of scores, four participants obtained thefollowing deviation scores: +1,-2,+5,and-10.

(a)Which score reflects the highest raw score?

+1reflects the highest raw score since the deviation is only a singleunit.

(b)Which score reflects the lowest raw score?

-10 reflects the lowest raw score since the deviation is 10 units

(c)Rank-order the deviation scores in terms of their frequency, startingwith the score with the lowest frequency.

-10,+5, -2, +1

21.Foofy says a deviation of +5isalways better than a deviation of -5.Why is she correct or incorrect?

Foofyis correct. This is because a negative deviation indicates an errorwhile a positive deviation indicates a residual. For me I believe itis easier to identify where the residual happens than to identify theerror. Therefore a deviation of +5 is always better than a deviationof -5.

20.Assume that the data in question 25 reflect a highly skewed intervalvariable.

(a)What statistics would you compute to summarize these conditions?

Median

(b)We compute them to be 14, 12, and 8.5, respectively. What conclusionabout the study should you draw?

Fromthe results it is clear that the scores is normally distributed

(c)What conclusion would you draw about the populations produced by thisexperiment?

Regardingthe results above it is viable to conclude that there are more that asingle population that will be produced basing off the collecteddata.

Chapter5

6.What do SX,sX,andoX havein common in terms of what they communicate?

Theyboth describe a particular type of standard deviation

(b)How do they differ in terms of their use?

Sxmeasures the sample standard deviation while σx is the populationstandard deviation, meaning all the data were used to calculate it.

11.Ina condition of an experiment, a researcher obtains the followingcreativityscores:32 1 0 7 4 8 6 6 4Interms of creativity scores, interpret the variability using thefollowing:

(a)therange,

Range= 8-0 = 8 . the scores have a very wide range which means thecreativity of scores represents the population very well.

(b)the variance,

Mean= 41/10 = 4.1

Sumof (x-xbar) = -1.1 – 2.1 – 3.1 – 4.1 + 2.9 – 0.1 + 3.9 + 1.9+1.9- 0.1 = 0

Sumof (x-xbar)2= 1.21 + 4.41 + 9.61 + 16.81 + 8.41 + 0.01 + 15.21 + 3.61 +3.61 +0.01 = 62.9

Variance= 62.9/10-1 = 62.9/9 = 6.989

(c)the standard deviation.

Standarddeviation = sqrt 6.989 = 2.6437 thisimplies that the data is not widely spread

14.Aspart of studying the relationship between mental and physical health,youobtainthe following heart rates:

7372 67 74 78 84 79 71 76 767981 75 80 78 76 78

Interms of differences in heart rates, interpret these data using thefollowing:(a)the range

=84 – 67 = 17 implying that there is no avery wide range between the heart rates and that does notrepresentative of the population.

(b)the variance

Mean= 1297/17 = 76.3

Sumof (x-xbar) = -3.3 – 4.3 – 9.3 – 2.3 + 1.7 + 7.7 +2.7 – 5.3 –0.3 – 0.3 + 2.7 +4.7 – 1.3 + 3.7 + 1.7 – 0.3 + 1.7 = 0

Sumof (x-xbar)2 =

10.89+18.49+86.49+5.29+2.89+59.29+7.29+28.09+0.09+0.09+7.29+22.09+1.69+13.69+2.89+0.09+2.89= 269.53

Variance= 269.53/ 16 = 16.85

(c)the standard deviation.

Standarddeviation = 4.104 this implies that the data is not widely spread

19.Consider the results of this experiment:

ConditionA Condition B Condition C

12 33 47

11 33 48

11 34 49

10 31 48

  1. What “measures” should you compute to summarize the experiment?

Means

  1. These are ratio scores. Compute the appropriate descriptive statistics and summarize the relationship in the sample data.

Meanof condition A = 44/4 = 11

Conditionb = 32.75

C= 48

  1. How consistent does it appear the participants were in each condition?

Theparticipants were very consistent because in each condition therespective results fall in the same range of deviation from the mean.

  1. Does this relationship account for much of the variance in the scores?

Thisrelationship does not account for much of the variance in the scores.A better way to do this would be to find the standard deviations ofeach.

22.Consider these ratio scores from an experiment:

Condition1 Condition 2 Condition 3

18 83

1311 9

9 65

  1. What should you do to summarize the experiment?

Calculatethe mean of each condition

  1. Summarize the relationship in the sample data.

Meanof condition 1 = 18+13+9 = 40/3 = 13.33

Meanof condition 2 = 8+11+6 = 23/3 = 7.67

Meanof condition 3 = 12+5 = 17/3 = 5.67

  1. How consistent were the scores in each condition?

Veryconsistent

23. Saythat you conducted the experiment in question 22 on the entirepopulation.

(a)Summarize the relationship that you’d expect to observe.

Samplemean is equal to the population mean according to the central limittheory. Therefore, I would expect the relationship to remain thesame.

(b)How consistently do you expect participants to behave in eachcondition?

Theparticipants in each condition will also behave the same as thesample.

24.Comparing the results in questions 19 and 22, which experimentproduced the stronger relationship? How do you know?

Theexperiment in question 22 produced a stronger relationship since therange of their means is not very wide.

Chapter6

  1. For the data,

9,5, 10, 7, 9, 10, 11, 8, 12, 7, 6, 9

Mean(Xbar) = 103/12 = 8.6

X

X – Xbar

(X – Xbar)2

9

0.4

0.16

5

-3.6

12.96

10

1.4

1.96

7

-1.6

2.56

9

0.4

0.16

10

1.4

1.96

11

2.4

5.76

8

-0.6

0.36

12

3.4

11.56

7

-1.6

2.56

6

-2.6

6.76

9

0.4

0.16

Totals

-0.2

46.92

Variance= 46.92/11 = 4.2655 S.d = Sqrt 4.2655 = 2.065

  1. Compute the z-score for the raw score of 10.

Z= (10 – 8.6)/ 2.065 = 0.678

  1. Compute the z-score for the raw score of 6.

Z= (6 – 8.6)/ 2.065 = -1.259

10.In an English class last semester, Foofy earned a 76(X= 85, SX=10).Her friend, Bubbles, in a different class, earned a 60(X=50, SX=4).Should Foofy be bragging about how much better she did? Why?

Foofyshould not brag since her score has much deviation from the meancompared to her friend who seems to have closely scored everything

13.For the data in question 13, find the raw scores that correspond tothe following:

(a)z= +1.22

Letx rep the raw score

Therefore1.22 = (x – 8.6)/2.065

1.22*2.065+ 8.6 = x

Theraw score X = 11.12

  1. z = -0.48.

-0.48= (X-8.6)/2.065

X= -0.48*2.065 + 8.6 = 7.61

15.Which z-scorein each of the following pairs corresponds to the lower raw score?

z=+1.0 or z=+2.3

z=+2.3corresponds to lower raw score

  1. z= –2.8 or z= –1.7

z=-2.8corresponds to lower raw score

(c)z=-.70 or z=-.20

z=-.70 corresponds to lower raw score

(d)z=0.0 or z=-2.0.

z=-2.0 corresponds to lower raw score

20.For an IQ test, we know the population u=100 andthe oX=16.We are interested in creating the sampling distribution when N=64.

(a)What does that sampling distribution of means show?

Thesampling distribution of means shows a distribution of sample meansof a statistic.

(b)What is the shape of the distribution of IQ means and the mean of thedistribution?

Theshape is a normal distribution

  1. Calculate o _X for this distribution.

Accordingto the central limit theory the mean of a sample distribution isequal to the population mean = 100

Thereforeo _X= 16/8 = 2

(d)What is your answer in part (c) called, and what does it indicate?

Itis called standard error and it indicates the range in which aconfidence interval in which the population means is likely to fall

  1. What is the relative frequency of sample means above 101.5?

=101.5/64 = 1.6