Merrill Lynch Securities and Health Care Retirement Inc. are two large employers in downtown Toledo, Ohio. They are considering jointly offering child care for their employees. As a part of the feasibility study, they wish to estimate the mean weekly child care cost of their employees. A sample of 10 employees who use child care reveals the following amounts spent last week.$107 $92 $97 $95 $105 $101 $91 $99 $95 $104(a) Develop a 99 percent confidence interval for the population mean. Interpret the results.

Step-by-step explanation:

We would determine the mean an standard deviation first

Mean = (107 + 92 + 97 + 95 + 105 + 101 + 91 + 99 + 95 + 104)/10 = 98.6

Standard deviation = √(summation(x - mean)/n

n = 10

Summation(x - mean) = (107 - 98.6)^2 + (92 - 98.6)^2 + (97 - 96.6)^2 + (95 - 98.6)^2 + (105 - 98.6)^2 + (101 - 98.6)^2 + (91 - 98.6)^2 + (99 - 98.6)^2 + (95 - 98.6)^2 + (104 - 98.6)^2 = 274

Standard deviation = √(274/10) = 5.23

The confidence interval is used to determine the range of values that could possibly contain a population parameter (population mean)

Confidence interval is written in the form,

(Point estimate - margin of error, Point estimate + margin of error)

The sample mean, x is the point estimate for the population mean.

We will use the t distribution because the sample size is small and the population standard deviation is not given.

Degree of freedom, df = 10 - 1 = 9

α = 1 - 0.99 = 0.01

z α/2 = 0.01/2 = 0.005

The area to the right of z 0.005 is 0.005 and the area to the left of z0.005 is 1 - 0.005 = 0.995. Using the t distribution table,

z = 3.25

Margin of error = z × s/√n

Where

s = sample standard Deviation = 5.23

Margin of error = 3.25 × 5.23/√10 = 5.38

Confidence interval = sample mean(point estimate) ± z × σ/√n

The lower boundary of the confidence interval is

98.6 - 5.38 = 93.22

The upper boundary of the confidence interval is

98.6 + 5.38 = 103.98

We estimate with 99% confidence that the mean weekly child care cost of their employees is between $93.22 and $103.98

Also,

99% of the confidence intervals constructed in this way would contain the true value for the population mean of mean weekly child care cost of their employees.

first one because it can't have a fraction in it

step-by-step explanation:

i cant answer this, theres no table. sorry man

step-by-step explanation: