A Construct a 90 confidence interval to estimate the populat

A) Construct a 90% confidence interval to estimate the population proportion with a sample proportion equal to .35 and a sample size equal to 120.

use the cumulative probabilities for the standard normal distribution table

lower limit _____ upper limit_____

B) Determine the margin of error for a confidence interval to estimate the population mean with n=32 and o=40 for the following confidence intervals.

91% =

96% =

99% =

Solution

a)
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
Mean(x)=42
Sample Size(n)=120
Sample proportion = x/n =0.35
Confidence Interval = [ 0.35 ±Z a/2 ( Sqrt ( 0.35*0.65) /120)]
= [ 0.35 - 1.64* Sqrt(0.002) , 0.35 + 1.64* Sqrt(0.002) ]
= [ 0.279,0.421]

b) AT 91% LOS
Margin of Error = Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
Mean(x)=18.6
Standard deviation( sd )=32
Sample Size(n)=40
Margin of Error = Z a/2 * 32/ Sqrt ( 40)
= 1.7 * (5.0596)
= 8.6014

AT 96% LOS
Margin of Error = Z a/2 * 32/ Sqrt ( 40)
= 2.05 * (5.0596)
= 10.3723

AT 99% LOS
Margin of Error = Z a/2 * 32/ Sqrt ( 40)
= 2.58 * (5.0596)
= 13.0539