Question

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.

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: 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.