A survey was conducted in 30 companies to study the percenta

A survey was conducted in 30 companies to study the percentage of engineering employees who contributed to some kind of community service. The survey gathered the following data:

76 81 77 82 80 85 60 80 79 82 70 88 85 80 79 83 75 87 78 80 84 72 75 90 84 82 77 75 86 80

Construct a 90% confidence interval for the mean participation rate for engineering employees

Solution

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)=79.733
Standard deviation( sd )=5.959
Sample Size(n)=30
Confidence Interval = [ 79.733 ± t a/2 ( 5.959/ Sqrt ( 30) ) ]
= [ 79.733 - 1.699 * (1.088) , 79.733 + 1.699 * (1.088) ]
= [ 77.885,81.581 ]