Question

The average selling price of a smartphone purchased by a random sample of 44 customers was ​$295. Assume the population standard deviation was ​$32. a. Construct a 90​% 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 90​% confidence interval has a lower limit of ​$ ? and an upper limit of ​$ ?

​(Round to the nearest cent as​ needed.)

b. What is the margin of error for this interval?

Answer 1

Solution

Given that,

= 295

=32

n = 44

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z/2 = Z0.05 = 1.645

Margin of error = E = Z/2* (/n)

= 1.645 * (32 / 44 )

= 7.93

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

- E < < + E

295- 7.93< < 295 +7.93

287.06 < < 302.93

The lower limit of ​$287

THE upper limit of ​$ 303