8.5.39) Determine the sample size n needed to construct a 99% confidence interva

Question

8.5.39) Determine the sample size n needed to construct a 99% confidence interval to estimate the population mean when =30 and the margin of error equals 5. n= (Round up to the nearest integer.)

8.5.41) Determine the sample size n needed to construct a 90% confidence interval to estimate the population proportion when p= 0.67 and the margin of error equals 5%. n= (Round up to the nearest integer.)

8.5.42) Determine the sample size n needed to construct a 90% confidence interval to estimate the population proportion when p=0.35 and the margin of error equals 7% n= (Round up to the nearest integer.)

8.6.50) Construct a 90% confidence interval to estimate the population mean using the data below. x=70 , =15 , n=30 , N=300 The 90% confidence interval for the population mean is ( , ) (Round to two decimal places as needed.)

Explanation / Answer

8.5.39.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.01% LOS is = 2.58 ( From Standard Normal Table )
Standard Deviation ( S.D) = 30
ME =5
n = ( 2.58*30/5) ^2
= (77.4/5 ) ^2
= 239.63 ~ 240      
8.5.40.
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.1 is = 1.645
Sample Proportion = 0.67
ME = 0.05
n = ( 1.645 / 0.05 )^2 * 0.67*0.33
= 239.321 ~ 240
8.5.41.
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.1 is = 1.645
Sample Proportion = 0.35
ME = 0.07
n = ( 1.645 / 0.07 )^2 * 0.35*0.65
= 125.637 ~ 126
8.5.42.
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=70
Standard deviation( sd )=15
Sample Size(n)=30
Confidence Interval = [ 70 ± Z a/2 ( 15/ Sqrt ( 30) ) ]
= [ 70 - 1.64 * (2.74) , 70 + 1.64 * (2.74) ]
= [ 65.51,74.49 ]

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