Question

A survey of 50 retail stores revealed that the average price of a microwave was $375, with a sample standard deviation of $20. Assuming the population is normally distributed, what is the 95% confidence interval to estimate the true cost of the microwave?

Answer 1

(369.4563 , 380.5437)

Step-by-step explanation:

Given that,

z_c = 1.96 at 95% confidence ⇒ Z- critical value

n = 50

x - bar = 375

SD = 20

Margin of error, E = ?

`E=(z_c* SD)/(√(n) )`

`E=(1.96* 20)/(√(50) )`

E = 5.5437

Therefore,

95% CI:

= (375 - 5.5437, 375 + 5.5437)

= (369.4563 , 380.5437)

Hence, the 95% confidence interval to estimate the true cost of the microwave is (369.4563 , 380.5437).