Question

A sample size of 120 had a sample mean and standard deviation of 100 and 15 respectively. Of thes...

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.

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Answer #1

Chi-square test of goodness of fit

Solution:

Here, we have to use chi square test for goodness of fit.

Null hypothesis: H0: Data follows normal distribution.

Alternative hypothesis: Ha: 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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