Question

Monthly rental rate data were collected from a large random sample of apartments in Manhattan. The mean rate was $3156.50 per month. The distribution of sample means was normal with a standard error of $92.801. Construct a 95% confidence interval to estimate the mean rate in the population of all Manhattan apartments.

[2974.610, 3338.390]

[3003.855, 3309.145]

[3138.311, 3174.689]

[3151.681, 3158.319]

Answer 1

Solution :

Given that,

Point estimate = sample mean =     = $3156.50

standard error SE=$92.801

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z / 2   = Z0.025 = 1.96   ( Using z table )

Margin of error = E =   Z / 2    *SE
=1.96 * 92.801

= 181.88996
At 95% confidence interval estimate of the population mean
is,

- E <   < + E

3156.50 - 181.88996 <   <3156.50 + 181.88996

2974.610 <   < 3338.390

( 2974.610 , 3338.390 )