Question

The average selling price of a smartphone purchased by a random sample of 34 customers was $302. Assume the population standard deviation was $30.
a. Construct a 95?% confidence interval to estimate the average selling price in the population with this sample.
b. What is the margin of error for this? interval?

a. The 95?% confidence interval has a lower limit of $__ and an upper limit of ?$__.
?(Round to the nearest cent as? needed.)
b. The margin of error is $__.
? (Round to the nearest cent as? needed.)

Answer 1

Solution :

Given that,

= 302

= 30

n = 34

(a) At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

(b) Margin of error = E = Z_/2* ( /n)

= 1.96 * (30 / 34)

= 10.08

Margin of error = $10.08

At 95% confidence interval estimate of the population mean is,

- E < < + E

302 - 10.08 < < 302 + 10.08

291.92 < < 312.08

(291.92 , 312.08)

Lower limit = $291.92

Upper limit = $312.08