Question

25) Suppose Ray's Restaurant wants to know how many people on an average day are in their restaurant. To find this out, Ray takes a random sample of 45 past days. He finds out that there was an average of 200 people per day with a sample standard deviation (3) of 36 people. Create a 90% Confidence Interval on the number of people for an average day in Ray's Restaurant: (4 points)

Answer 1

Solution :

Given that,

t_{$\alpha$ /2,df} = 1.680

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 1.680 * (36 / $\sqrt$ 45)

Margin of error = E = 9

The 90% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

200 - 9 < $\mu$ < 200 + 9

191 < $\mu$ < 209

(191 , 209)