The Human Resources Department of a large manufacturing company wants to estimat

Question

The Human Resources Department of a large manufacturing company wants to estimate the proportion of employees who have received CPR training. Based on a simple random sample of 150 employees, it was found that 48 of them have CPR training. It is desired to construct a 98% confidence interval for the proportion of all employees who have CPR training.

Answer the following questions. Show your work or your calculator commands.

A) Find the point estimate,p^, of the population proportion.

B) Find the critical value of the confidence interval.

C) Find the margin of error of the confidence interval.

D) Construct a 98% confidence interval for the proportion of all employees who have CPR training.

E) If the Human Resources Department wants to keep the margin of error at 0.04 for the 98% confidence interval, what size of sample should be taken?

Explanation / Answer

a)
People have CPR training(x)=48
Sample Size(n)=150
Sample proportion = x/n =0.32
b)
crtical value at 98% confidence is 2.33
c)
Margin of Error = Z a/2 Sqrt(p*(1-p)/n))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=48
Sample Size(n)=150
Sample proportion =0.32
Margin of Error = Z a/2 * ( Sqrt ( (0.32*0.68) /150) )
= 2.33* Sqrt(0.0015)
=0.0887
d)
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval

Confidence Interval = [ 0.32 ±Z a/2 ( Sqrt ( 0.32*0.68) /150)]
= [ 0.32 - 2.33* Sqrt(0.0015) , 0.32 + 2.33* Sqrt(0.0015) ]
= [ 0.2313,0.4087]

e)
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.02 is = 2.33
Samle Proportion = 0.32
ME = 0.04
n = ( 2.33 / 0.04 )^2 * 0.32*0.68
= 738.3304 ~ 739

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