The maker of potato chips uses an automated packaging machine to pack its 20-ounce bag of chips. at the end of every shift 18 bags are selected at random and tested to see if the equipment needs to be readjusted. after one shift, a sample of 18 bags yielded the following data. mean = 20.45 s = .80 n = 18. if we were to conduct a test to see if the sample estimate is different from the company’s expected weight of 20 oz. at an alpha level = .05, what is the conclusion for this test?

Let be the population mean .

By considering the given information in the question , we have

Here, is two-tailed , so the test is a two-tailed test.

Also, population standard deviation is unknown , so we perform  a two-tailed t-test.

For sample size : n= 18

Sample mean :

Sample standard deviation : s=0.80

Test statistic :

Two-tailed Critical values for and degree of freedom df=n-1=17  

Decision : Since the calculated t-value(2.39) does not lie between the critical values -2.11 and 2.11.

So we reject the null hypothesis .

Conclusion : We have enough evidence at 0.05 significance level to support the claim that the sample estimate is different from the company’s expected weight of 20 oz.

6.5

step-by-step explanation:

you said it was in the question

~courtney