Consider a population with a known standard deviation of 26.8. In order to compu

Question

Consider a population with a known standard deviation of 26.8. In order to compute an interval estimate for the population mean, a sample of 64 observations is drawn. Use Table 1.

Compute the margin of error at a 95% 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 95% confidence level based on a larger sample of 225 observations.(Round your intermediate calculations to 4 decimal places. Round "z" value and final answer to 2 decimal places.)

Which of the two margins of error will lead to a wider confidence interval?

Consider a population with a known standard deviation of 26.8. In order to compute an interval estimate for the population mean, a sample of 64 observations is drawn. Use Table 1.

Explanation / Answer

a)
Yes
b)
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
Standard deviation( sd )=26.8
Sample Size(n)=64
Margin of Error = Z a/2 * 26.8/ Sqrt ( 64)
= 1.96 * (3.35)
= 6.566
c)
Sample Size(n)=225
Margin of Error = Z a/2 * 26.8/ Sqrt ( 225)
= 1.96 * (1.787)
= 3.502
d)
With the sample size 64 gives the wider confidence

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