QuestionDetails: a recent study of 25 employees of XYZ company showed the mean(a

Question

QuestionDetails: a recent study of 25 employees of XYZ company showed the mean(average) of the distance they traveled to work was 14.3 miles. thestander deviation of the sample mean was 2 miles. find the 95%confidence of the true mean.


(b) if the manager wanted to be sure that most of the employeeswould not be late, how much time should she suggest each employeeallow, assuming they travel 30 miles per hour
I really need help with part B.
QuestionDetails: a recent study of 25 employees of XYZ company showed the mean(average) of the distance they traveled to work was 14.3 miles. thestander deviation of the sample mean was 2 miles. find the 95%confidence of the true mean.


(b) if the manager wanted to be sure that most of the employeeswould not be late, how much time should she suggest each employeeallow, assuming they travel 30 miles per hour
I really need help with part B.
QuestionDetails: a recent study of 25 employees of XYZ company showed the mean(average) of the distance they traveled to work was 14.3 miles. thestander deviation of the sample mean was 2 miles. find the 95%confidence of the true mean.


(b) if the manager wanted to be sure that most of the employeeswould not be late, how much time should she suggest each employeeallow, assuming they travel 30 miles per hour
I really need help with part B.

Explanation / Answer

= 14.3 standard deviation, = 2 n = 25 standard error, e = /n = 2 / 5 = 0.4 for 95 %, z = 1.65 hence confidence interval, CI = ± ze = 14.3± (1.65 x 0.4) = (13.64,14.96)
b) the employees travel 30 miles per hour lets consider the right end of the confidence intervalcalculated above. that means 14.96 miles will be covered in (1/30)x14.96 hr= 0.499 hrs.= 29.92 minutes ˜ 30 minutes hence if the manager wanted to be sure that most of theemployees would not be late, she should suggest each employeeto be allowed 30 minutes

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