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.

The degrees of freedom are given by:

And the p value would be given by:

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

represent the sample size

represent the confidence level  

represent the sample variance obtained

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:

Alternative hypothesis:

Calculate the statistic  

The statistic would be given by:

Repalcing we got:

The degrees of freedom are given by:

And the p value would be given by:

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

robert puts 2 in each row

to plot points   for x just start at the origin (middle) and go to the left however many times it say(if the number is a positive) if the number is a negative then go right however many times it says if it is 0 go nowhere

now keep you pencil there and go up however many times it says on y(if it is positive) if it is negative then go down however many times it says. if it says 0 go nowhere.

you have just plotted a point! now when you finish plotting all of them connect them with a ruler

plot each point as (-3, 7), (0,4), (3,1), (8,-4)   in the parenthesis on the left of the comma is x and the right is y.