Exercise 803 Algorithmic A simple random sample of 60 items

{Exercise 8.03 (Algorithmic)}

A simple random sample of 60 items resulted in a sample mean of 88. The population standard deviation is 16.

a. Compute the 95% confidence interval for the population mean (to 1 decimal).

(  ,   )

b. Assume that the same sample mean was obtained from a sample of 120 items. Provide a 95% confidence interval for the population mean (to 2 decimals).

(  ,   )

Solution

a)
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)=88
Standard deviation( sd )=16
Sample Size(n)=60
Confidence Interval = [ 88 ± Z a/2 ( 16/ Sqrt ( 60) ) ]
= [ 88 - 1.96 * (2.066) , 88 + 1.96 * (2.066) ]
= [ 83.951,92.049 ] ~ [ 85.1,90.9 ]

b)
Sample Size(n)=120
Confidence Interval = [ 88 ± Z a/2 ( 16/ Sqrt ( 120) ) ]
= [ 88 - 1.96 * (1.461) , 88 + 1.96 * (1.461) ]
= [ 85.137,90.863 ] ~ [ 85.14,90.86 ]