We are given;
Sample size, n = 11
Sample mean, x' = $553
Sample standard deviation, s = $179
90% confidence interval
We need to calculate;
the margin of error, M
Formula for the margin of error, M for a confidence interval:

t is the critical value for a 2 tailed test at a 10% level of significance
degree of freedom = sample size - 1
df = 11 - 1 = 10 and the level of significance α = 10%
The value of t from a t distribution table is 1.8125
Thus;
the margin of error is, M = $97.82

The 90% confidence interval for the population mean is:
