Instructor's Manual Student Assistants[1][1]

8/3/2019 Instructor's Manual Student Assistants[1][1]

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Instructors Manual

Business Research Methods Pre-Msc: Statistics

(2008/09)

Week 46 & 47:

Chapter 1


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Chapter 2


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Chapter 4


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Chapter 6


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Chapter 7


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Chapter 8


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Week 48:

Chapter 9


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Chapter 10


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Chapter 11


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Week 49:

Chapter 12


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Chapter 13


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Week 50

Chapter 15


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SPSS-exercises

15.47 Analyze > Descriptive Statistics > Crosstabs (Row Variable: Education; Column Variable:

Section). Press the button Statistics and select Chi-square, then OK.

2 = 86.615, p=0.000. Conclusion: There is sufficient evidence to infer that educational level

affects the way adults read the newspaper.

15.48 For this exercise it is better to use Excel. Make a table with the observed (which are given)

and expected frequencies (p*the total number of draws=1/49*312=6.37). Compute the chi-square


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statistic and compare the obtained value with the critical value (df = 48; =0.05). In Excel, thecorresponding p-value can be computed by using the function: CHIDIST.

2 = 38.22, p=0.8427. Conclusion: There is not enough evidence to infer that the numbers were

not generated randomly.


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Chapter 19


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SPSS-exercises

19.79 Conduct a Spearman Rank Correlation (Analyze > Correlate > Bivariate and select

Spearman). Rho = 0.574 and p Nonparametric Tests > 2 Independent Samples> (Test Variable: Section; Grouping

Variable: Group).

Z=2.65 and p


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Analyze > Nonparametric Tests > K Independent Samples (Test Variable: Binding; GroupingVariable: Group).


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Week 51:

Section 4.4


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Chapter 16


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SPSS-exercises

16.81 Analyze > Regression > Linear (Dependent: Customer; Independent: Ads).


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a. xy *36.2192.296 += .

b. On average each additional ad generates 21.36 (22) customers.

c. Coefficient b1 is not significant (t=1.495, p>0.05). There is not enough evidence to conclude

that the larger the number of ads, the larger the number of customers.

d. R square = 0.085. There is a weak linear relationship between the number of ads and the

number of customers.

e. The linear relationship is too weak for the model to produce predictions.


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Chapter 17


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SPSS-exercises

17.40 Analyze > Regression > Linear (Dependent: Yield; Independents: Fertilizer, Water).


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a. 21 31.014.001.164 xxy ++= . For each additional unit of fertilizer crop yield increases by

0.14 (holding the amount of water constant). For each additional unit of water crop yield

increases on average by 0.31 (holding the fertilizer constant).

b. b1 is not significant (t-value = 1.717, p>0.05). There is not enough evidence to infer that there

is a linear relationship between crop yield and the amount of fertilizer.c. b2 is significant (t-value = 4.637, p Regression > Linear), but indicate that you want to

save the standardized and unstandardized residuals and the unstandardized predictions. Make ahistogram of the (standardized) residuals (Graphs > Regression > Linear > Plots > Standardized

Residuals Plots > select Histogram). Conclusion the errors appear to be normal.


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Plot the (unstandardized) residuals against the predicted values (Graphs > Legacy Dialogs >

Scatter/Dot. Here seems to be a problem.

f. Choose Analyze > Regression > Linear > Save, and select besides unstandardized

predictions also individual prediction interval. The first row of your dataset give the

upper- (349.3) and lowerband (69.2) of your prediction (209.3).

Instructor's Manual Student Assistants[1][1]

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