STAT217
Worksheet #5 Solutions
Question 1
a)
Ho:
response does not depend on city
Ha: response depends on city
Degrees of freedom = 2 x 4 = 8. Reject Ho if c2 > 15.507
To get the test statistic, we need the row and
column totals:
|
|
Strongly Disagree |
Disagree |
Neutral |
Agree |
Strongly Agree |
Total |
|
Vancouver |
6 |
20 |
40 |
20 |
14 |
100 |
|
Calgary |
12 |
18 |
24 |
16 |
12 |
82 |
|
Toronto |
12 |
18 |
62 |
16 |
12 |
120 |
|
Total |
30 |
56 |
126 |
52 |
38 |
302 |
From there we can get the expected values:
|
|
Strongly Disagree |
Disagree |
Neutral |
Agree |
Strongly Agree |
|
Vancouver |
9.9338 |
18.5430 |
41.7219 |
17.2185 |
12.5828 |
|
Calgary |
8.1457 |
15.2053 |
34.2119 |
14.1192 |
10.3179 |
|
Toronto |
11.9205 |
22.2517 |
50.0662 |
20.6623 |
15.0993 |
And then the terms for the test statistic:
|
|
Strongly Disagree |
Disagree |
Neutral |
Agree |
Strongly Agree |
|
Vancouver |
1.5578 |
0.1145 |
0.0711 |
0.4493 |
0.1596 |
|
Calgary |
1.8237 |
0.5137 |
3.0482 |
0.2505 |
0.2742 |
|
Toronto |
0.0005 |
0.8124 |
2.8445 |
1.0520 |
0.6362 |
When we add the terms for the test statistic,
we get 13.6082.
Do not reject Ho.
Conclude the response does not depend on what
city the person is from.
b)
Cramers
V = sqrt(13.6082/(2x302)) = 0.1501 = 15.01%
c)
The
10% critical value is 13.362. We reject Ho and conclude that the response
depends on what city the person is from.
Question 2
a)
When
we work out the expected values for the 20+ category, this is what we get:
|
working |
semi-retired |
retired |
|
1.33 |
3.62 |
5.05 |
As we see, 2 of the 3 expected values are less
than 5. Thus, we need to collapse this category with the 10 to 19 category.
b)
Here
is the new crosstab of observed values:
|
< 10 |
10 or more |
Total |
|
|
Working |
10 |
4 |
14 |
|
Semi-retired |
13 |
25 |
38 |
|
Retired |
15 |
38 |
53 |
|
Total |
38 |
67 |
105 |
Ho: hours worked does not depend on work status
Ha: hours worked does depend on work status
Degrees of freedom = 2 x 1 = 2; Reject Ho if
test statistic > 5.991.
Here is the crosstab of expected values:
|
5.067 |
8.933 |
|
13.752 |
24.248 |
|
19.181 |
33.819 |
Test statistic = 4.803 + 2.724 + 0.041 + 0.023
+ 0.911 + 0.517 = 9.019
Reject Ho.
Conclude hours worked depends on work status
c)
Cramers
V = sqrt(9.019/105) = 0.2931 = 29.31%
Question 3
Ho: data is
normal
Ha: data is
not normal
Reject Ho
if test statistic > 0.319.
Here is the
completed table:
|
X |
Z |
F(z) |
S(z) |
S'(z) |
D |
|
0 |
-1.1667 |
0.1217 |
0.1667 |
0 |
0.1217 |
|
2 |
-0.4298 |
0.3337 |
0.5000 |
0.1667 |
0.1670 |
|
2 |
-0.4298 |
0.3337 |
0.5000 |
0.1667 |
0.1670 |
|
3 |
-0.0614 |
0.4755 |
0.6667 |
0.5 |
0.1912 |
|
4 |
0.3070 |
0.6206 |
0.8333 |
0.6667 |
0.2127 |
|
8 |
1.7808 |
0.9625 |
1.0000 |
0.8333 |
0.1292 |
Test
statistic = 0.2127.
Do not
reject Ho and conclude the data is normally distributed.
From the textbook
Section
10-2
Question 9 (page
563)
Ho: digits
are uniformly distributed
Ha: digits
are not uniformly distributed
Reject Ho
if c2 > 16.919 (based on 9 df)
c2 = (8 10)2/10 + (8 10)2/10
+ (12 10)2/10 + (11 10)2/10 + (10 10)2/10
+ (8 10)2/10 + (9 10)2/10 + (8 10)2/10 +
(12 10)2/10 + (14 10)2/10 = 0.4 + 0.4 + 0.4 + 0.1 + 0
+ 0.4 + 0.1 + 0.4 + 0.4 + 1.6 = 4.2
Do not
reject Ho. Conclude the digits are uniformly distributed.
Question 10
(page 563)
Ho: digits
are uniformly distributed
Ha: digits
are not uniformly distributed
Reject Ho
if c2 > 16.919 (based on 9 df)
c2 = (0 10)2/10 + (17 10)2/10
+ (17 10)2/10 + (1 10)2/10 + (17 10)2/10
+ (16 10)2/10 + (0 10)2/10 + (16 10)2/10
+ (16 10)2/10 + (0 10)2/10 = 10 + 4.9 + 4.9 + 8.1 +
4.9 + 3.6 + 10 + 3.6 + 3.6 + 10 = 63.6
Reject Ho.
Conclude the digits are not uniformly distributed.
Section
10-3
Question 6
(page 579)
Ho: error
rate does not depend on price
Ha: error
rate does depend on price
Df = (2-1)(3-1) = 2; reject Ho if c2 > 5.991
c2 = (20 13.8)2/13.8 + (7 13.2)2/13.2
+ (15 22.5)2/22.5 + (29 21.5)2/21.5 + (384 382.7)2/382.7
+ (364 365.3)2/365.3 = 2.7855 + 2.9121 + 2.5 + 2.6163 + 0.0044 +
0.0046 = 10.8229
Reject Ho.
Conclude the error rates are not the same for regular-priced items as for
advertised-special items. Based on the expected values for the over-priced
items, it does appear that advertised-priced items are overpriced more. If I
believed a store was overcharging me on advertised-priced items, Id be
shopping somewhere else. Note that Cramers V = sqrt(10.8229/819) = 11.5%
which indicates that over/undercharging depends on the price 11.5% of the time.
Section
10-4
Question 1 (page
592)
Ho: the data is normally distributed
Ha: it is not
Reject Ho if test statistic > 0.285
The mean of the data is 278.125 and the standard deviation is 436.8551. This spreadsheet summarizes the results:
|
Value |
Z
score |
F(z) |
S(z) |
S'(z) |
Max diff |
|
50 |
-0.52 |
0.3015 |
0.1250 |
0 |
0.3015 |
|
75 |
-0.46 |
0.3228 |
0.2500 |
0.125 |
0.1978 |
|
100 |
-0.41 |
0.3409 |
0.3750 |
0.25 |
0.0909 |
|
120 |
-0.36 |
0.3594 |
0.5000 |
0.375 |
0.1406 |
|
140 |
-0.32 |
0.3745 |
0.6250 |
0.5 |
0.2505 |
|
150 |
-0.29 |
0.3859 |
0.7500 |
0.625 |
0.3641 |
|
240 |
-0.09 |
0.4641 |
0.8750 |
0.75 |
0.4109 |
|
1350 |
2.45 |
0.9929 |
1.0000 |
0.875 |
0.1179 |
|
278.125 |
mean |
|
|
test stat |
0.4109 |
|
436.8551 |
stdev |
|
|
|
|
The test statistic is 0.4109. Reject Ho and conclude the data is not normally distributed.
Question 2
(page 592)
Ho/Ha are the same.
Since n=7, reject Ho if test statistic > 0.3
The mean is now 125 and the standard deviation is 61.7117. Here are the results:
|
Value |
Z
score |
F(z) |
S(z) |
S'(z) |
Max diff |
|
50 |
-1.22 |
0.1112 |
0.1429 |
0 |
0.1112 |
|
75 |
-0.81 |
0.209 |
0.2857 |
0.1429 |
0.0767 |
|
100 |
-0.41 |
0.3409 |
0.4286 |
0.2857 |
0.0877 |
|
120 |
-0.08 |
0.4681 |
0.5714 |
0.4286 |
0.1033 |
|
140 |
0.24 |
0.5948 |
0.7143 |
0.5714 |
0.1195 |
|
150 |
0.41 |
0.6591 |
0.8571 |
0.7143 |
0.1980 |
|
240 |
1.86 |
0.9686 |
1.0000 |
0.8571 |
0.1115 |
|
125 |
mean |
|
|
test stat |
0.1980 |
|
61.7117 |
stdev |
|
|
|
|
The test statistic is 0.198. Do not reject Ho and conclude the data is normally distributed.