A market researcher wants to find the 99 Confidence Interval

A market researcher wants to find the 99% Confidence Interval for the average number of hours people spend online each day. She obtained a small sample of 18 people and found the sample mean was 92.5 minutes with a sample standard deviation of 15.3. Find the margin of error for the 99% Confidence Interval for the population mean.

a. 2.575 b. 10.5 c. 63.2 d. 9.3

Solution

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)=13
Standard deviation( sd )=15.3
Sample Size(n)=18
Margin of Error = t a/2 * 15.3/ Sqrt ( 18)
= 2.898 * (3.606)
= 10.451 ~ 10.5 [ANSWER]