A statistics practitlioner took a randoml sample of 54 observations form a popu

Sinead Mcgee

Sinead Mcgee

Answered question

2021-10-02

A statistics practitlioner took a randoml sample of 54 observations form a population whose standard deviation is 29 and computed the sample mean to be 97.
Note: For each confidence interval, enter your answer in the forem (LCL, UCL). You must include the parentheses and the comma between the confidence limits.
A) Estimate the population mean with 95% confidence.
B) Estimate the population mean with 90% confidence.
C) Estimate the population mean with 99% confidence.

Answer & Explanation

krolaniaN

krolaniaN

Skilled2021-10-03Added 86 answers

Setp 1
Given:
σ=29
n=54

x=97
a) From the Standard Normal Table, the value of z for 95% level is 1.96
The 95% confidence interval for the population mean is obtained as below:
Sample statistic ±zSE=x±zσn
=97±(1.96×2954)
97±7.7349
=(89.2651, 104.7349)
Thus, the 95% confidence interval for the population mean is (89.2651, 104.7349).
Step 2
b) From the Standard Normal Table, the value of z for 90% level is 1.645.
The 90% confidence interval for the population mean is obtained as below:
Sample statistic ±zSE=x±zσn
=97±(1.645×2954)
=97±6.4918
=(90.5082, 103.4918)
Thus, the 90% confidence interval for the population mean is (90.5082, 103.4918).
Step 3
c) From the Standard Normal Table, the value of z for 99% level is 2.576
The 99% confidence interval for the population mean is obtained as below:
Sample statistic ±zSE=x±zσn
=97±(2.576×2954)
=97±10.1659
=(86.8341, 107.1659)
Thus, the 99% confidence interval for the population mean is (86.8341, 107.1659).

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