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A recent study of 28 employees of XYZ company showed that the mean of the distance they traveled to work was 14.3 miles. The standard deviation of the sample mean was 2 miles.a. Find the 95% confidence interval of the true mean.b. If a manager wanted to be sure that most of her/his employees would not be late, how much time would she/he suggest they allow for the commute if the average speed were 30 miles per hour

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nuhulawal20

Answer:

- At 95% confidence interval, the true mean is ( 13.5245 < μ < 15.0755 )

- the time allowed will be  0.50 hours or 30 minutes

Step-by-step explanation:

Given the data in the question;

sample size; n = 28

mean; x" = 14.3 miles

standard deviation; S = 2 miles.

degree of freedom DF = n - 1 = 28 - 1 = 27

confidence interval = 95%

level of significance = 1 - 95% = 1 - 0.95 = 0.05

so

[tex]t_{\alpha /2, df[/tex] = [tex]t_{0.025, df=27[/tex] = 2.0518

Hence, we have;

x" + [tex]t_{\alpha /2, df[/tex]( S/√n ) = 14.3 + 2.0518( 2/√28 )

= 14.3 + 0.7755

= 15.0755     { Upper Limit }

Also,

x" - [tex]t_{\alpha /2, df[/tex]( S/√n ) = 14.3 - 2.0518( 2/√28 )

= 14.3 - 0.7755

= 13.5245    { Lower Limit }

Therefore, at 95% confidence interval, the true mean is ( 13.5245 < μ < 15.0755 )

b)

If a manager wants to be sure that the employees are not late, then he/she should consider the upper bound of the confidence interval as the permissible distance range.

Now given that the average speed were 30 miles per hour

suggested time will be;

t = Upper limit / speed

t = 15.0755 / 30

t = 0.50 hours or 30 minutes

Therefore, the time allowed will be  0.50 hours or 30 minutes