Consider a population with a known standard deviation of 129

Consider a population with a known standard deviation of 12.9. In order to compute an interval estimate for the population mean, a sample of 39 observations is drawn.

Use Table 1. === http://lectures.mhhe.com/connect/0077639472/Table/table1.jpg

Compute the margin of error at a 90% confidence level. (Round your intermediate calculations to 4 decimal places. Round \"z\" value and final answer to 2 decimal places.)

Compute the margin of error at a 90% confidence level based on a larger sample of 325 observations.(Round your intermediate calculations to 4 decimal places. Round \"z\" value and final answer to 2 decimal places.)

Solution

a)
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)=125.6
Standard deviation( sd )=12.9
Sample Size(n)=39
Margin of Error = Z a/2 * 12.9/ Sqrt ( 39)
= 1.64 * (2.0657)
= 3.3877
b)
Sample Size(n)=325
Margin of Error = Z a/2 * 12.9/ Sqrt ( 325)
= 1.64 * (0.7156)
= 1.1735