Question

The length, in words, of the essays written for a contest are normally distributed with a population standard deviation of 442 words and an unknown population mean. If a random sample of 24 essays is taken and results in a sample mean of 1330 words, find a 99% confidence interval for the population mean.

Answer 1

Answer: $(1076.71\ ,1583.29)$

Step-by-step explanation:

Given : Level of significance : $1-\alpha:0.99$

Then , significance level : $\alpha: 1-0.99=0.01$

Since , sample size : $n=24$ , which is small sample (n<30) so the test applied here is a t-test.

Degree of freedom= $n-1=24-1=23$

Using t-distribution table , Critical value : $\text{t-score}=t_{n-1, \alpha/2}=2.807336$

Sample mean : $\overline{x}=1330$

Standard deviation: $\sigma: 442$

The confidence interval for population mean is given by :-

$\overline{x}\pm t_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

$1330\pm(2.807336)\dfrac{442}{\sqrt{24}}\approx1330\pm253.29=(1076.71\ ,1583.29)$

Hence,  a  99% confidence interval for the population mean.=$(1076.71\ ,1583.29)$