Question

In 36 randomly selected seawater samples, the mean sodium chloride concentration was 23 cubic centimeters per cubic meter. Assume the population standard deviation is 6.7 cubic centimeters per cubic meter.
Construct the 90% and 95% confidenceintervals for the population mean.

Answer 1

Answer: 90% confidence interval for population mean =(21.163, 24.837)

95% confidence interval for population mean =(20.811, 25.189)

Step-by-step explanation:

Confidence interval for population mean ($\mu$) is given by :-

$\overline{x}\pm z \dfrac{\sigma}{\sqrt{n}}$ , where n= sample size, z= critical z-value , $\overline{x}$= sample mean $\sigma$= population standard deviation.

Given : n= 36 ,  $\overline{x}=23$ cubic centimeters per cubic meter.

$\sigma=6.7$cubic centimeters per cubic meter

For 90% confidence level , z= 1.645 [by z-table]

Then, required interval for population mean ($\mu$) would be :

$23\pm (1.645) \dfrac{6.7}{\sqrt{36}}$

$=23\pm 1.837$

$=(23-1.837 , 23+ 1.837)=(21.163, 24.837)$

For 95% confidence level , z= 1.96 [by z-table]

Then, required interval for population mean ($\mu$) would be :

$23\pm (1.96) \dfrac{6.7}{\sqrt{36}}$

$=23\pm 2.189$

$=(23-2.189 , 23+ 2.189)=(20.811, 25.189)$

Hence, 90% confidence interval for population mean =(21.163, 24.837)

95% confidence interval for population mean =(20.811, 25.189)