A man regularly time himself as he runs 3 miles a day From l

A man regularly time himself as he runs 3 miles a day. From long experience, he knows that his times are normally distributed, with an overall standard deviation of 3.2 minutes. In a random sample of 115 runs, his average time is 21.4 minutes. Find a 99% confidence interval for the mean of all his running times. What is the margin of error for this confidence interval?

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)=21.4
Standard deviation( sd )=3.2
Sample Size(n)=115
Confidence Interval = [ 21.4 ± Z a/2 ( 3.2/ Sqrt ( 115) ) ]
= [ 21.4 - 2.58 * (0.298) , 21.4 + 2.58 * (0.298) ]
= [ 20.63,22.17 ]

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 )=3.2
Sample Size(n)=115
Margin of Error = Z a/2 * 3.2/ Sqrt ( 115)
= 2.576 * (0.298)
= 0.7687