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 99% confidence interval to estimate the true cost of the microwave

Answer 1

The 99% C.I is ( 367.703, 382.297 )

Explanation:

Given that:

sample mean
`\overline x` = 375

Population standard deviation
`\sigma` = 20

Sample size n = 50

The level of significance
`\alpha` = 1 - C.I = 1 - 0.99 = 0.01

From z tables:

The critical value =
`Z_(\alpha /2)` =
`Z_(0.005)` = 2.58

The margin of error E =
`Z_(\alpha /2)`
`* (\sigma )/(√(n))`

The margin of error E =
`2.58 * (20)/(√(50))`

The margin of error E = 7.297

The limits of 99% C.I are expressed as:

Lower limit =
`\overline x - E` = 375 - 7.297

Lower limit = 367.703

Upper limit =
`\overline x + E` = 375 + 7.297

Upper limit = 382.297

The 99% C.I is
`\overline x \pm E` =
`375 \pm 7.297`

= ( 367.703, 382.297 )