Question

For an aptitude test for quality-control technicians in an electronics firm, history shows scores to be normally distributed with a variance of 2500. For 20 applications who took the test, the sample standard deviation was 68. Is there enough evidence to reject the assumption that the variance of scores is 2500. Use significance level of 0.05.

Answer 1

Answer:

$\chi^2 =\frac{20-1}{2500} 68^2 =35.14$

The degrees of freedom are given by:

$df = n-1=20-1=19$

And the p value would be given by:

$p_v =2*P(\chi^2 >35.14)=0.0268$

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true variance for this case is different from 2500

Step-by-step explanation:

Information given

$n=20$ represent the sample size

$\alpha=0.05$ represent the confidence level  

$s^2 =68^2=4624$ represent the sample variance obtained

$\sigma^2_0 =2500$ represent the value that we want to test

Null and alternative hypothesis

We want to test if the true variance is 2500, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 = 2500$

Alternative hypothesis: $\sigma^2 \neq 2500$

Calculate the statistic  

The statistic would be given by:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

Repalcing we got:

$\chi^2 =\frac{20-1}{2500} 68^2 =35.14$

The degrees of freedom are given by:

$df = n-1=20-1=19$

And the p value would be given by:

$p_v =2*P(\chi^2 >35.14)=0.0268$

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true variance for this case is different from 2500