Use the t -distribution to find a confidence interval for a mean mu given the re

Question

Use the t -distribution to find a confidence interval for a mean mu given the relevant sample results. Give the best point estimate for mu, the margin of error, and the confidence interval. Assume the results come from a random sample from a population that is approximately normally distributed. A 90% confidence interval for mu using the sample results bar x = 132.6, s = 52.7, and n = 50 Round your answer for the point estimate to one decimal place, and your answers for the margin of error and the confidence interval to two decimal places. point estimate = 132.6 margin of error = 12.26 The 90% confidence interval is 120.34 to 144.86

Explanation / Answer

a.
Margin of Error = t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
Mean(x)=132.6
Standard deviation( sd )=52.7
Sample Size(n)=50
Margin of Error = t a/2 * 52.7/ Sqrt ( 50)
= 1.677 * (7.453)
= 12.499

b.
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=132.6
Standard deviation( sd )=52.7
Sample Size(n)=50
Confidence Interval = [ 132.6 ± t a/2 ( 52.7/ Sqrt ( 50) ) ]
= [ 132.6 - 1.677 * (7.453) , 132.6 + 1.677 * (7.453) ]
= [ 120.101,145.099 ]

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