Question

Help with a hypothesis testing problem.

As a result of a long timing of the Assembly time of the node by different assemblers, it is found that the variance of this time, σ₀² = 2 min² (NB sigma-nought squared = 2 min squared).  The results of the 20 observations of the work of a beginner are as follows

xi	56	58	60	62	64
ni	1	4	10	3	2

Can it be considered at the level of significance α = 0,05 that the workers work rhythmically (in the sense that the variance of the time spent is not significantly different from the dispersion of time of other collectors)?

(Taking H₀ : σ² = 2 and the alternative hypothesis H_a : σ² ≠ 2 perform a hypothesis test for this problem.)

Answer 1

Solution:

Here, we have to use the Chi square test for the population variance.

H₀: σ^2 = 2 versus H_a: σ^2 ≠ 2

This is a two tailed test.

We are given level of significance = α = 0.05

Sample size = n = 20

Degrees of freedom = df = n – 1 = 20 – 1 = 19

Sample standard deviation = S = 1.9974

(Calculated from given data)

Lower critical value = 8.9065

Upper critical value = 32.8523

(by using Chi square table or excel)

The test statistic formula is given as below:

Chi square = (n – 1)*S^2/ σ^2

Chi square = (20 - 1)* 1.9974^2 / 2

Chi square = 37.90126

P-value = 0.0061

(by using Chi square table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that the variance of the time spent by workers is significantly different from the dispersion of time of other collectors.