Question

A manufacturer believes that the variability of one of its processes is o?y=0.0005 millimeter?. A sample is taken from the process (sample size n = 6) and the sample variance is calculated to be s?y= 0.0020. Assuming that the measurements follow a normal distribution, evaluate the manufacturer's claim assuming an a value of 0.02. a) What statistic will you use to solve the problem? b) What is the value of the statistic? c) Do you agree with the claim of the manufacturer? Answer Yes or No.

Answer 1

The null and the alternate hypothesis re

$H_{o}: \sigma^2 = 0.0005$

$H_{a}: \sigma^2 \neq 0.0005$

a) We will use the Chi-squared statistics to solve the problem.

b) The value of the chi-squared statistics is given by

$\chi^2 = \frac{(n-1)s^2}{\sigma^2}$

x2 (6 – 1) * 0.0020 0.0005 20

c) The p-value corresponding to the test statistics of 20 is given by

= 2* PGX2 > 20)

The significance level is 0.02.

Since the p-value is less than the significance level, hence we will reject the null hypothesis.

Answer - No, we don't agree with the claim of the manufacturer.

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