Question

Let us consider a sample of 35 continuous measurements:

12.5	11.9	13.2	17.4	13.5	15.7
11.9	13.7	16.5	12.4	11.6	17.2
15.6	14.9	12.8	17.5	15.7	11.6
14.9	15.2	13.7	13.3	12.6	11.7
17.3	15.6	14.6	13.6	11.6	12.6
15.3	13.8	16.4	15.8	17.4

Test whether this sample is from a normal population or not. Use a 5% level of significance.

I've already calculated the mean to be 13.31 and the standard deviation to be 1.94 but I'm not sure how to find the z-intervals.

Answer 1

Solution

Solution is based on the theory of Chi-sqaure Test for Goodness of fit

Final answers are given below. Back-up Theory and Details of calculations follow at the end.

The given sample is from a normal population. Answer

[Z value = (x-value – Mean)/Standard deviation.]

Back-up Theory and Details of calculations

Construction of Frequency Distribution and Computation of Mean and Standard Deviation.

Class Boundary	Frequency
LCB	UCB
11.5	12.5	8
12.5	13.5	6
13.5	14.5	4
14.5	15.5	5
15.5	16.5	7
16.5	17.5	5
	Total	35
Mean	14.3143
SD	1.9374

Goodness of Fit

Let Oi and Ei be respectively the observed and expected frequencies of the ith class, i = 1 to k, k being the number of classes given.

Hypotheses:

Null: H₀: Observed frequencies are in accordance with Normal Distribution

Vs

Alternative H_A: H₀ is false

Test Statistic:

χ2 = ∑[i = 1,k]{(Oi - Ei)²/Ei},

Under H₀, Ei’s follow N(14.3143, 1.9374).

Details of calculations

Class No (i)	UCB (Ui)	Oi	Zi	Φ(Zi)	pi	Ei	ChiSq
1	12.5	8	-0.9312	0.1759	0.1759	6.1557	0.5526
2	13.5	6	-0.4177	0.3381	0.1622	5.6774	0.0183
3	14.5	4	0.0958	0.5382	0.2001	7.0028	1.2876
4	15.5	5	0.6093	0.7288	0.1907	6.6737	0.4198
5	16.5	7	1.1228	0.8692	0.1404	4.9140	0.8855
6	17.5	5	5.0000	1.0000	0.1308	4.5764	0.0392
Total	35			1.0000	35.0000	3.2030

Chisquare Cal	3.2030
DF	4	Alpha	0.05
Chisquare Crit	9.487729
p-value		0.170816

Distribution, Significance Level, α, Critical Value, p-value

Under H₀, χ² ~ χ²_{k – s}, Chi-square distribution with degrees of freedom = k – s,where k = number of classes and s =number of parameters estimated.

p-value = P(χ²_{k – s} > χ²_cal)

Given significance level = α , critical value = χ²_crit = upper α% of χ²_{k - s, α}

Critical value and p-value obtained using Excel Function: Statistical CHIINV and CHIDIST are as shown in the above table.

Decision

Since, χ2_cal < χ2_crit, or equivalently, since p-value > α, H₀ is accepted

Conclusion

There is sufficient statistical evidence to conclude that the given data conform to Normal distribution.

DONE