STAT217 Tutorial sheet #5 Solutions
Question 1
Ho: the distribution is uniform
Ha: the distribution is not uniform
Since there are 4 categories, the degrees are 3. Reject Ho if c2 > 7.81
To compute the test statistic, since the distribution is uniform, each category should have 25 observations.
c2 = (28 25)2/25 + (30 25)2/25 + (23 25)2/25 + (19 25)2/25 = 2.96
Do not reject Ho.
Conclude the distribution is uniform.
Question 2
Ho: the distribution is binomial with n=7 and p=0.24
Ha: it is not
To get the expected value we multiply P(x) by 500 for each value of x:
|
x |
500P(x) |
|
0 |
73.25 |
|
1 |
161.85 |
|
2 |
153.35 |
|
3 |
80.7 |
|
4 |
25.5 |
|
5 |
4.85 |
|
6 |
0.5 |
|
7 |
0 |
You will note the sum of the second column is 500. Since the expected value of the last three values is less than 5, we combine them to get 5.35. The sum of the observed values for these categories is 4+0+0 = 4.
Since we now have 6 categories instead of 8, the degrees of freedom are 5. Reject Ho if c2 > 11.0705
c2 = (85-73.25)2/73.25 + (159-161.85)2/161.85 + + (4-5.35)2/5.35 = 4.7763.
Do not reject Ho.
Conclude the data follows a binomial distribution with n=7 and p=0.24.
Question 3
Ho: the distribution is Poisson with mean = 3
Ha: it is not
To get the expected value we multiply P(x) by 500 for each value of x:
|
x |
500P(x) |
|
0 |
24.9 |
|
1 |
74.7 |
|
2 |
112 |
|
3 |
112 |
|
4 |
84 |
|
5 |
50.4 |
|
6 |
25.2 |
|
7 |
10.8 |
|
8 |
4.05 |
Since the expected value of 8 is less than 5, we stop. We note that the sum of 500P(0) through 500P(7) (i.e. the expected values for 0 through 7) is 494; that means the sum of the remaining expected values is 500 494 = 6 for x=8 onward. The sum of the observed values for the last category is then 1+2 = 3.
Since we have 9 categories, the degrees of freedom is 8. Reject Ho if c2 > 15.5073.
c2 = (26-24.9)2/24.9 + (84-74.7)2/74.7 + + (3-6)2/6 = 5.3892
Do not reject Ho.
Conclude the data follows a Poisson distribution with mean = 3.
Question 4
a) 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.
b) 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.
Question 5
a) Ho: home status does not depend on gender
Ha: it does
degrees of freedom = 3; Reject Ho if c2 > 7.8147
First we need the totals:
|
|
Male |
Female |
Total |
|
Stable home |
3658 |
8099 |
11757 |
|
Home with problems |
138 |
293 |
431 |
|
Transient home situation |
357 |
708 |
1065 |
|
Homeless |
2450 |
3047 |
5497 |
|
Total |
6603 |
12147 |
18750 |
Next, the expected values:
|
|
Male |
Female |
|
Stable home |
4140.3451 |
7616.6549 |
|
Home with problems |
151.7810 |
279.2190 |
|
Transient home situation |
375.0504 |
689.9496 |
|
Homeless |
1935.8235 |
3561.1765 |
Finally, the terms for the test statistic:
|
|
Male |
Female |
|
Stable home |
56.1926 |
30.5458 |
|
Home with problems |
1.2512 |
0.6802 |
|
Transient home situation |
0.8687 |
0.4722 |
|
Homeless |
136.5711 |
74.2388 |
The sum of the terms gives c2 = 300.8206.
Reject Ho and conclude that home status depends on gender.
b) Cramers V = sqrt(300.8206/18750) = 0.1267 = 12.67%