Question

A company produces thousands of plastic toy parts every day. So that the toys can be assembled properly a certain tab thickness x has to meet quality control specifications. The mean of the population of x has to be within a certain range and the standard deviation of population of x has to be less than a certain value. A quality control inspector took a random sample of 18 parts and measured x.He computed a sample mean of 8.2mm for x and a sample standard deviation of 0.024mm. Having found the mean was in spec next he has to test sample variance to decide whether to reject the null hypothesis that the population variance of x is less than or equal to the quality tolerance 0.0002 mm2 using an alpha og 0.05. He assumes the population mean follows a normal distribution?

A) draw a diagram that shows reject/fail to reject regions for this problem?

b)What is the statistic used

c) What is the test statistic value computed from the sample statistics given

D) Since the level of significance alpha is 0.05 state the critical value(s) which would be use relevant to the question. State your conclusion about the null hypothesis.

Answer 1

Solution:

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

Null hypothesis: H₀: the population variance of x is less than or equal to the quality tolerance 0.0002 mm^2.

Alternative hypothesis: H_a: the population variance of x is greater than to the quality tolerance 0.0002 mm^2.

H₀: σ² ≤ 0.0002 versus H_a: σ² > 0.0002

This is an upper tailed test.

We are given

Level of significance = α = 0.05

Sample size = n = 18

Degrees of freedom = df = n – 1 = 18 – 1 = 17

Sample standard deviation = S = 0.024

Upper critical value = 27.5871

(by using Chi square table)

A) Draw a diagram that shows reject/fail to reject regions for this problem?

The required diagram is given as below:

b) What is the statistic used

The statistic used for this test is Chi square statistic for the testing for population variance.

c) What is the test statistic value computed from the sample statistics given

The test statistic formula is given as below:

Chi square = (n – 1)*S^2/ σ²

Chi square = (18 - 1)*0.024^2/0.0002

Chi square = 48.96

Test statistic value = 48.96

D) Since the level of significance alpha is 0.05 state the critical value(s) which would be use relevant to the question. State your conclusion about the null hypothesis.

We have

Test statistic value = 48.96

Upper critical value = 27.5871

Test statistic value > Upper critical value

So, we reject the null hypothesis

There is insufficient evidence to conclude that the population variance of x is less than or equal to the quality tolerance 0.0002 mm^2.