Answer:
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.
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