. Construct a 98% confidence interval for the average monthly apartment rent using a random sample...

. Construct a 98% confidence interval for the average monthly apartment rent using a random sample of 32 apartments with a mean of $375 and a standard deviation of 32.5.

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Expert Solution

Solution :

Given that,

Point estimate = sample mean = = $375

Population standard deviation = = $32.5

Sample size = n =32

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z_/2 = Z_0.01 = 2.326 ( Using z table )

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

= 2.326 * ( 32.5 / $\sqrt$ 32 )

E= 13.3634
At 98% confidence interval estimate of the population mean
is,

- E < < + E

375 - 13.3634 < < 375 + 13.3634

361.6366< < 388.3634

( 361.6366, 388.3634 )

orchestra answered 1 year ago

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