The tread depth of the right front tire was measured for 10

The tread depth of the right front tire was measured for 10 randomly stopped automobiles. The mean was 0.30 inch, and the standard deviation was 0.05 inch. Find the 95% confidence interval of the mean depth. Assume that the variable is approximately normally distributed

Solution

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)=0.3
Standard deviation( sd )=0.05
Sample Size(n)=10
Confidence Interval = [ 0.3 ± t a/2 ( 0.05/ Sqrt ( 10) ) ]
= [ 0.3 - 2.262 * (0.016) , 0.3 + 2.262 * (0.016) ]
= [ 0.264,0.336 ]