Assume that a sample is used to estimate a population mean . Use
the given confidence level and sample data to find the margin of
error. Assume that the sample is a simple random sample and the
population has a normal distribution. Round your answer to one more
decimal place than the sample standard deviation. (please show
work) 
Homework Answers
Degrees of freedom = n-1
= 51 - 1
= 50
t critical value at 0.05 significance level for 50 df is 2.009
Margin of error = t * S / sqrt(n)
= 2.009 * 202 / sqrt(51)
= 56.8
Margin of error = 56.8 (Rounded one more decimal places than sample standard deviation)
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