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Review Question Answers
Note: x’ is used as the symbol for the mean
sqrt stands for square root
1. mean = 6.38
st dev = 1.61
2. Confidence interval (two tailed):
 = 0.05
df = n-1 = 4
t,df = 2.78
Interval = 6.38 +/- 2.78 * 1.61/sqrt(5)
= 6.38 +/- 2.00
[4.38 to 8.38]
3. c (because the value 28.2 is an outlier which may influence the mean)
4. Overall mean = (5 * 53.8) + (4 * 47.4) + (5 * 52.2) + (4 * 43.8) / 18 = 49.7 mm
5. c (two independent sample t-test)
6.  = 0.85, e = 2.71828, x = 1
P(X) = 0.851 * 2.71828-0.85
1!
= 0.36
7. a (one sample KS test using Poisson’s)
8. d (errors terms must be homoscedastic – equal variance)
9. a (when many overcast days, we expect more cases of depression)
10. b
11. b (from table, b0 = 26.473, b1 = -0.492, equation is Y = 26.473 – 0.492X)
12. a (df are number of independents = 1 and n-2 = 33)
13. c (cannot calculate correlation with nominal data)
14. a (two tailed hypotheses test only for significant correlation, not positive or
negative)
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15. c
16. Translate into Z score
Z = ($30,000 - $38,500)/$12,643
= -0.67
Look up probability on Z table:
P(0 to –0.67) = 0.2486
P(-0.67 to end) = 0.5 – 0.2486 = 0.2514
Probability of income less than $30,000 = 0.25
17. Translate into Z score
Z = ($50,000 - $38,500)/$12,643
= 0.91
Z = ($60,000 - $38,500)/$12,643
= 1.70
Look up probability on Z table:
P(0 to 0.91) = 0.3186
P(0 to 1.70) = 0.4554
P(0.91 to 1.70) = 0.4554 – 0.3186 = 0.1368
Probability of income between $50,000 and $60,000 = 0.14
18.
Test: use one sample difference of means test because comparing sample value
to known value
Assumptions: data ($) are ratio, population distribution assumed to be normal,
population mean ($36,750) is known
Hypotheses: (non-directional)
H0: sample mean equals known value (x’ = )
Ha: sample mean greater or less than known value (x’  )
Prob distribution: use t distribution
Significance level: use  = 0.05 (standard 95% confidence)
Critical value: at  = 0.05 and df (n-1) = 40-1 = 39, tc for two tailed test = 2.02
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Test statistic:
x’ (1999) = $38,500,  (1995) = $36,750, s = $12,643, n = 40
t* = 38,500 – 36,750
12,643/sqrt(40)
= 0.875
Decision:
reject H0 if t* > tc
t* (0.875) is less than tc (2.02) so cannot reject H0
Inference: we infer with 95% confidence that there is no significant difference
between the 1999 mean income and the 1995 mean income in Victoria.
19. d (use formula for small sample because n=22)
20. e (use formula for sample containing individual observations)
21. c (paired t-test because each sample contains the same individuals,
directional hypotheses to determine if internet use has increased)
22.
Test: use one sample chi-square test with uniform expected frequencies to see if
frequencies (number of students with access to computer) are different between
the schools.
Assumptions: schools are nominal categories, categories are mutually exclusive,
frequencies (number of students) are absolute counts, expected frequencies for
uniform distribution are: 290/5 = 58 (all expected frequencies greater than 5).
Hypotheses:
H0: there is no difference between the observed and expected frequencies
Ha: there is a significant difference between observed and expected frequencies
Probability distribution: use 2 distribution
Significance level: use  =0.05 (standard 95% confidence)
Critical value: at  = 0.05 and df = k-1 = 4, 2c = 9.49
Test statistic
School
Observed
Eagle Bay
82
Elm Tree
43
Glen Valley
58
King George
36
Ridge Crest
71
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Expected
58
58
58
58
58
(O-E)2/E
9.93
3.88
0.00
8.34
2.91
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Total
290
25.07
2* = 25.07
Decision:
Reject H0 if 2* > 2c
2* (25.07) is greater than 2c (9.49) so we reject H0
Inference: We infer with 95% confidence that there is a difference between the
observed and expected (uniform) frequencies.
23. d (test will only tell you if there are differences, you cannot make any causal
inferences)
24. p = allergy/total children = 210/608 = 0.35
25. Confidence interval (two tailed):
= 0.05, /2=0.025
p=0.35
Z/2 = 1.96
Interval = 0.35 +/- 1.96 * sqrt(0.35 *(1-0.35))
608
= 0.35 +/- 0.038
[0.31 to 0.39]
26. P(allergies given industrial city) = 89/209 = 0.43
27. P(rural area given allergies) = 43/210 = 0.20
28. a (chi-square because data are nominal, df = (k-1)*(l-1) = 1 * 2 = 2)
29.
Answer
d
b
f
c
g
a
e
Inferential test
One sample difference of means
Regression
Two or more sample Chi-square
Spearman’s
1 sample chi square
1-sample K-S
2 sample difference of means
30. c
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31. Problem: this sample may not represent the true population of people who
shop at the store. Because this is a non-probability sample, the retailer will not
be able to use any inferential tests on the data. However, he could use
descriptive statistics on his dataset if he wants.
32. c (directional because they are testing if the mean of sample A is greater
than the mean of sample B)
33. a (at  = 0.01 and df (n1+n2-2) = 15+15 –2 = 28, tc for one tailed test is 2.47).
34. Because the data are unreliable, the researcher should use a nonparametric test for multiple samples – the chi-square test which uses frequencies
and categories instead of actual data values.
35. With unreliable data, the researcher should use a non-parametric test for
differences between two samples – the two sample KS test (which also uses
rankings). The researcher could also use the 2 or more sample Chi-square test
but it would be better to downgrade to an ordinal test, rather than a nominal test
(less information will be lost).
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