Question

Monthly expenses for an advertising agency were obtained for the previous year. The sample of 12 expenses showed a mean of $196,000, with a standard deviation of $45,000. The office manager is budgeting for monthly expenses, and would like to know if it is reasonable to expect expenses to be below $250,000. Construct a 95% confidence interval to justify if $250,000 is reasonable.

Answer 1

We know that:

Sample mean, xbar= 196000

Sample standard deviation, s= 45000

Sample size, n= 12

T statistic, t_11,0.025= 2.201

Thus, interval

= xbar - t * s / sqrt(n) , xbar + t * s / sqrt(n)

= 196000 - 2.201 * 45000/ sqrt(12) , 196000 + 2.201 * 45000/ sqrt(12)

= 167408.2, 224591.8

Thus, since $250,000 is outside this interval, we know that this is not a reasonable estimate to make.