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A population is normal with a variance of 9. Suppose you wish to estimate the population mean. Find the sample size needed to assure with 90% confidence that the samnple mean will not differ from the population by mnore than 2 units.



A sample of 25 different payroll departments found that the employees worked an average of 310.3 days a year with a standard deviation of 23.8 days. What is the 90% confidence interval for the average days worked in all payroll departments? Assume the number of days worked is normally distrributed.

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Dec 15, 2010
Statistics – Sample Size
by: Staff

The question:


A population is normal with a variance of 9.


Suppose you wish to estimate the population mean.


Find the sample size needed to assure with 90% confidence that the sample mean will not differ from the population by more than 2 units.


A sample of 25 different payroll departments found that the employees worked an average of 310.3 days a year with a standard deviation of 23.8 days.

What is the 90% confidence interval for the average days worked in all payroll departments?

Assume the number of days worked is normally distributed.

The answer:

Part I of your question:

Find the sample size needed to assure with 90% confidence that the sample mean will not differ from the population by more than 2 units.


Actual Population Mean = your estimated value of mean ± margin of error you will accept

The margin of error you will accept is + or – 2 units

Margin of Error formula = (Z value)*(Standard Deviation)/( square root of sample size)

Since + or – 2 units is the maximum margin of error you will accept, then the Margin of Error must be less than or equal to 2 units

2 ≥ Margin of Error

2 ≥ (Z value)*(Standard Deviation)/(square root of sample size)


N = sample size

Confidence level, z* value
90% 1.645
95% 1.96
98% 2.33
99% 2.58

The Z value for a 90% confidence limit = 1.645

The standard deviation for your problem is the square root of the variance
= sqrt(9) = 3

Substituting all known values:

2 ≥ (1.645)*(3)/(sqrt N)

Solving for N

N ≥ [(1.645)*(3)/(2)]^2

N ≥ 6.08855625

The final answer to Part I is: the sample size needed to assure 90% confidence must be greater than or equal to 6.08855625:


Part II of your question:

What is the 90% confidence interval for the average days worked in all payroll departments?

True Average of Days Worked = your estimated value of days worked ± margin of error

True Average of Days Worked = 310.3 days ± margin of error in days

Margin of Error = (Z value)*(Standard Deviation)/( square root of sample size)

The Z value for a 90% confidence limit = 1.645
Standard Deviation = 23.8
Sample Size = 25


Margin of Error = ± (1.645)*(23.8)/( square root of 25)

Margin of Error = ± (1.645)*(23.8)/( 5)

Margin of Error = ± 7.8302 days


True Average of Days Worked = 310.3 days ± 7.8302 days



Thanks for writing.


Staff
www.solving-math-problems.com


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