Following table shows the calculations:
O | E=205/12 | (O-E)^2/E | |
9 | 17.083 | 3.824555933 | |
11 | 17.083 | 2.166065035 | |
13 | 17.083 | 0.975875959 | |
18 | 17.083 | 0.049223731 | |
19 | 17.083 | 0.215119651 | |
26 | 17.083 | 4.654503834 | |
22 | 17.083 | 1.415260142 | |
29 | 17.083 | 8.313228883 | |
26 | 17.083 | 4.654503834 | |
15 | 17.083 | 0.253988702 | |
11 | 17.083 | 2.166065035 | |
6 | 17.083 | 7.190358192 | |
Total | 205 | 204.996 | 35.87874893 |
Following is the test statistics:
The degree of freedom: df=n-1=12-1=11
The p-value using excel function "=CHIDIST(35.879,11)" is: 0.000
Since p-value is less than 0.05 so we reject the null hypothesis.
There is sufficient evidence....
----------------------------
(c)
The confidence interval is 0.030 <p < 0122
(d)
Correct option is B.
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