A manufacturer of floor mops would like her product to last 700 hours, on the average. She hopes the mean number of hours is not a lot less than 700 (people will not continue to buy her product) or a lot more than 700 (people would seldom have to buy the product). However, the manufacturer has reason to believe the mean may have changed. A random sample of 48 items shows that sample mean = 675 and sample standard deviation = 77 hours. Use the significance level α=.05. What is the appropriate test to perform? Regarding the hypothesis testing in question 1, what is the rejection region? Regarding the hypothesis testing in question 1, what is the p-value of the test? Round your answer to four decimal places Regarding the hypothesis testing in question 1, what is your conlusion?

As the population standard deviation is unknown and the sample size is more than 30, it is appropriate to use one sample t test for this hypothesis testing

Testing if the mean is changed from 700 levels or not

Null and alternate hypothesis are 700

Degree of freedom,

n-1

= 48-1

= 47

Using alpha level 0.05 and t critical table, we found the following two critical values

t critical values= -2.012 and 2.012

So, the rejection reason is t statistic less than -2.012 or t statistic more than -2.012

Now,

Population means = 700 hours

Sample mean = 675 hours

n = 48

So,

Test statistic= (mean-alternate hypothesis)/(s/√n)

= (675-700)/ (77/√48)

=-25/11.1140

= -2.249

Degree of freedom = n-1

= 48-1

= 47

With t distribution table for -2.249 and 47 degrees of freedom obtain p value of 0.0292

So, pvalue= 0.0292

Reject null hypothesis since the p value is less than alpha level

So, we conclude that the mean has changed from 700 hours

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