Question

Question Help 8.1.1 Assuming the population of interest is approximately normally distributed, construct a 96% confidence interval estimate for the population mean given the values below. X = 16.9 54.3 ns12 The 95% confidence interval for the population mean is from to (Round to two decimal places as needed. Use ascending order.)

Answer 1

Answer:

Given Data

= 16.9

T

s = 4.3

n = 12

A 95% Confidence level

Level of significance

$\alpha=1-0.95$

= 0.05

Critical value

$t_{\alpha/2,n-1}$

$t_{0.05/2,11}$

$t_{0.025,11}$

$t_{\alpha/2 ,11}=2.200985$

Using t table

The 95% confidence interval for $\mu$ is

$\bar{X}-\left ( t_{\alpha/2} \right )\left ( \frac{s}{\sqrt{n}} \right )<\mu<\bar{X}+\left ( t_{\alpha/2} \right )\left ( \frac{s}{\sqrt{n}} \right )$$=16.9-(2.200985)\left ( \frac{4.3}{\sqrt{12}} \right )<\mu<16.9+(2.200985)\left ( \frac{4.3}{\sqrt{12}} \right )$$=16.9-2.7321 < \mu <16.9+2.7321$

$=14.1679< \mu <19.6321$

The 95% confidence interval for the population mean is from 14.1679 to 19.6321

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