A simple random sample of 60 items resulted in a sample mean

A simple random sample of 60 items resulted in a sample mean of 79. The population standard deviation is 14. a. 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)=79
Standard deviation( sd )=14
Sample Size(n)=60
Confidence Interval = [ 79 ± Z a/2 ( 14/ Sqrt ( 60) ) ]
= [ 79 - 1.96 * (1.807) , 79 + 1.96 * (1.807) ]
= [ 75.46,82.54 ]
b)
Sample Size(n)=120
Confidence Interval = [ 79 ± Z a/2 ( 14/ Sqrt ( 120) ) ]
= [ 79 - 1.96 * (1.278) , 79 + 1.96 * (1.278) ]
= [ 76.5,81.5 ]