A student randomly surveyed n 36 other students and found t

A student randomly surveyed n = 36 other students and found that the mean amount of money spent on texts was \\bar{x} x = $121.60. If the standard deviation of the population was \\sigma = $6.36, find the 90% confidence interval of the mean.

Solution

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)=121.6
Standard deviation( sd )=6.36
Sample Size(n)=36
Confidence Interval = [ 121.6 ± Z a/2 ( 6.36/ Sqrt ( 36) ) ]
= [ 121.6 - 1.64 * (1.06) , 121.6 + 1.64 * (1.06) ]
= [ 119.862,123.338 ]