A sample size of 120 had a sample mean and standard deviation of 100 and 15 respectively.
Of these 120 data values, 3 were less than 70, 18 were between 70 and 85, 30 between 85 and 100, 35 between 100 and 115, 32 were between 115 and 130 and 2 were greater than 130. Test the hypothesis that the sample distribution was normal.
Chi-square test of goodness of fit
Solution:
Here, we have to use chi square test for goodness of fit.
Null hypothesis: H_{0}: Data follows normal distribution.
Alternative hypothesis: H_{a}: Data do not follow normal distribution.
We assume level of significance = α = 0.05
Test statistic formula is given as below:
Chi square = ∑[(O – E)^2/E]
Where, O is observed frequencies and E is expected frequencies.
We are given
N = 6
Degrees of freedom = df = N – 1 = 6 – 1 = 5
α = 0.05
Critical value = 11.07049775
(by using Chi square table or excel)
Calculation tables for test statistic are given as below:
Class |
O |
Prob. |
E |
(O - E)^2 |
(O - E)^2/E |
less than 70 |
3 |
0.02275 |
2.730016 |
0.07289145 |
0.02670001 |
Between 70 and 85 |
18 |
0.135905 |
16.30861 |
2.860784443 |
0.17541554 |
Between 85 and 100 |
30 |
0.341345 |
40.96137 |
120.1516219 |
2.933291131 |
Between 100 and 115 |
35 |
0.341345 |
40.96137 |
35.53792665 |
0.867596154 |
Between 115 and 130 |
32 |
0.135905 |
16.30861 |
246.2195746 |
15.09751625 |
Greater than 130 |
2 |
0.02275 |
2.730016 |
0.532923118 |
0.195208801 |
Total |
120 |
1 |
120 |
19.29572789 |
(Corresponding probabilities are calculated by using z-table.)
Chi square = ∑[(O – E)^2/E] = 19.29572789
P-value = 0.001692928
(By using Chi square table or excel)
P-value < α = 0.05
So, we reject the null hypothesis
There is insufficient evidence to conclude that Data follows normal distribution.
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