Answer to Question #263037 in Statistics and Probability for angrl

Question

The manager of a fast-food chain wants to know the average amount the customers spend in their store. Ten customers were asked and the following data were collected; 150, 200, 180, 150, 250, 100, 200, 300, 100, 200 (data are in pesos) with a standard deviation of 59.19 pesos. Find the interval estimate of the average amount spent by the customers at α = 0.05.

Answer 1

$\bar x=\frac{150=200+180+150+250+100+200+300+100+200}{10}=\frac{1830}{10}=183.$

Since the population standard deviation 59.19 is known, the critical z-value $z_{0.025}=1.96$

should be used to evaluate 95% confidence interval.

$95\%CI=(183-1.96\frac{59.19}{\sqrt{10}},183+1.96\frac{59.19}{\sqrt{10}})=$

$=(146.31,219.69).$