10 football fans were randomly selected and asked how much m

10 football fans were randomly selected and asked how much money they

spend on concessions at a single game. The sample produced a mean of $16.00 and standard

deviation of $3.35. Construct a 99% condence interval for the mean.

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)=16
Standard deviation( sd )=3.35
Sample Size(n)=10
Confidence Interval = [ 16 ± t a/2 ( 3.35/ Sqrt ( 10) ) ]
= [ 16 - 3.2498 * (1.059) , 16 + 3.2498 * (1.059) ]
= [ 12.557,19.443 ]