Question

the mean of a sample size n=35 is 1860. the standard deviation of the sample is 102 and the population is normally distributed. construct a 99% confidence interval estimate of the mean of the population.

Answer 1

Answer: (1812.967, 1907.033)

Step-by-step explanation:

The confidence interval for population mean is given by :-

$\overline{x}\pm t_{df,\ \alpha} \dfrac{s}{\sqrt{n}}$

, where $\overline{x}$ = Sample mean

n= Sample size.

s = Sample standard deviation

$t_{df,\ \alpha}$ = Critical t-value for degree of freedom(n-1).

As per given , we have

n= 35

$\overline{x}=1860$

s=102

Significance level : $\alpha=0.01$

By t- distribution table , for degree of freedom 43 and $\alpha=0.01$ , we have

$t_{df,\ \alpha}=t_{34,\ 0.005} =2.728$

Substitute all the values in the above formula , we will get

$1860\pm (2.728)\dfrac{102}{\sqrt{35}}$

$=1860\pm (2.728)(17.2411)$

$=1860\pm47.034$

$=(1860-47.033\ 1860+47.033)= (1812.967,\ 1907.033)$

Hence, the  99% confidence interval estimate of the mean of the population is  (1812.967, 1907.033).