To lessen the financial burden of child care, a company is considering to reimburse its employees...

To lessen the financial burden of child care, a company is considering to reimburse its employees for child care expenses incurred while on the job. A random sample of 24 employees indicates the average weekly child care expense is $147.50 with a sample standard deviation of $46.75. Assuming these expenses are normally distributed, calculate the margin of error that will result when estimating the mean weekly child care expense for all employees at the company with a confidence level of 99%.

Do not round intermediate calculations. Round your final answer to 2 decimal places. Omit the "$" sign.

Margin of error =
dollars

Solutions

Expert Solution

solution

Given that,

= $147.50

s =$46.75

n = 24

Degrees of freedom = df = n - 1 =24 - 1 = 23

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2 df = t0.005,23 =2.807 ( using student t table)

Margin of error = E = t/2,df * (s / $\sqrt$ n)

= 2.807 * ( 46.75/ $\sqrt$ 24)

E=26.79

Margin of error = E =$26.79

orchestra answered 1 year ago

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