77 students were asked to choose 3 of the integers 11, 12, 13, .., 30 completely
ID: 3254118 • Letter: 7
Question
77 students were asked to choose 3 of the integers 11, 12, 13, .., 30 completely arbitrarily. The amazing result was as follows. If the selection were completely random, the following hypotheses should be true. (a) The 20 numbers are equally likely. (b) The 10 even numbers together are as likely as the 10 odd numbers together. (c) The 6 prime numbers together have probability 0.3 and the 14 other numbers together have probability 0.7. Test these hypotheses, using alpha = 5%. Design further experiments that illustrate the difficulties of random selection. Check your generator experimentally by imitating results of n trials of rolling a fair die, with a convenient n (e).g., 60 or 300 or the like). Do this many times andExplanation / Answer
Answer:
a).
Ho: The 20 numbers are equally likely
H1: The 20 numbers are not equally likely
Goodness of Fit Test
observed
expected
O - E
(O - E)² / E
11
11.550
-0.550
0.026
10
11.550
-1.550
0.208
20
11.550
8.450
6.182
8
11.550
-3.550
1.091
13
11.550
1.450
0.182
9
11.550
-2.550
0.563
21
11.550
9.450
7.732
9
11.550
-2.550
0.563
16
11.550
4.450
1.715
8
11.550
-3.550
1.091
12
11.550
0.450
0.018
8
11.550
-3.550
1.091
15
11.550
3.450
1.031
10
11.550
-1.550
0.208
10
11.550
-1.550
0.208
9
11.550
-2.550
0.563
12
11.550
0.450
0.018
8
11.550
-3.550
1.091
13
11.550
1.450
0.182
9
11.550
-2.550
0.563
Total
231
231.000
24.325
24.325
chi-square
19
df
.1840
p-value
Calculated chi square =24.325
DF=20-1=19
Chi square value at 0.05 level =30.14
Calculated chi square =24.325 < 30.14 the table value.
The null hypothesis is not rejected.
We conclude that the 20 numbers are equally likely.
b).
Ho: The 10 odd numbers are equally likely as even numbers
H1: The 10 odd numbers are not equally likely as even numbers
Goodness of Fit Test
observed
expected
O - E
(O - E)² / E
Odd
143
115.500
27.500
6.548
even
88
115.500
-27.500
6.548
Total
231
231.000
13.095
13.095
chi-square
1
df
.0003
p-value
Calculated chi square =13.095
DF=1
Chi square value at 0.05 level =3.841
Calculated chi square =13.095 >3.841 the table value.
The null hypothesis is rejected.
We conclude that the 10 odd numbers are not equally likely as even numbers.
c).
prime number
f
11
11
13
20
17
21
19
16
23
15
29
13
Total
96
Ho: The prime numbers with p=0.3 and non prime numbers with p=0.7.
H1: The numbers are not by given probability.
Goodness of Fit Test
observed
expected
O - E
(O - E)² / E
Prime
96
69.300
26.700
10.287
Non prime
135
161.700
-26.700
4.409
Total
231
231.000
0.000
14.696
14.696
chi-square
1
df
.0001
p-value
Calculated chi square =14.696
DF=1
Chi square value at 0.05 level =3.841
Calculated chi square =14.696 >3.841 the table value.
The null hypothesis is rejected.
We conclude that the numbers are not by given probability.
Goodness of Fit Test
observed
expected
O - E
(O - E)² / E
11
11.550
-0.550
0.026
10
11.550
-1.550
0.208
20
11.550
8.450
6.182
8
11.550
-3.550
1.091
13
11.550
1.450
0.182
9
11.550
-2.550
0.563
21
11.550
9.450
7.732
9
11.550
-2.550
0.563
16
11.550
4.450
1.715
8
11.550
-3.550
1.091
12
11.550
0.450
0.018
8
11.550
-3.550
1.091
15
11.550
3.450
1.031
10
11.550
-1.550
0.208
10
11.550
-1.550
0.208
9
11.550
-2.550
0.563
12
11.550
0.450
0.018
8
11.550
-3.550
1.091
13
11.550
1.450
0.182
9
11.550
-2.550
0.563
Total
231
231.000
24.325
24.325
chi-square
19
df
.1840
p-value
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