The monthly incomes for12 randomly selected people, each with a bachelor\'s degr

Question

The monthly incomes for12 randomly selected people, each with a bachelor's degree ineconomics, are shown on the right. 4450.55
4596.76
4366.29
4455.22
4151.13
3727.39
4283.39
4527.16
4407.14
3946.98
4023.91
4221.43

Complete parts (a) through (c) below.

Assume the population is normally distributed.

(a) Find the sample mean.

x overbar=

(Round to one decimal place as needed.)

(b) Find the sample standard deviation.

s=

(Round to one decimal place as needed.)

(c) Construct a 99% confidence interval for the population mean .

A 99% confidence interval for the population mean is

The monthly incomes for12 randomly selected people, each with a bachelor's degree ineconomics, are shown on the right. 4450.55
4596.76
4366.29
4455.22
4151.13
3727.39
4283.39
4527.16
4407.14
3946.98
4023.91
4221.43

Complete parts (a) through (c) below.

Assume the population is normally distributed.

Explanation / Answer

solution:-
(a)sample mean
x over bar = 4263.1

(b)sample standard deviation
s = 260.0

(c)99% confidence with df = 11 is t = 3.106
confidence interval formula
=> x +/- t * s / sqrt(12)
=> 4263.1 +/- 3.106 * 260.0 / sqrt(12)
=> (4029.9775 , 4496.2225)

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