Question

A simple random sample with n = 58 provided a sample mean of 24.5 and a sample standard deviation of 4.4. (Round your answers to one decimal place.) (a) Develop a 90% confidence interval for the population mean. to

Answer 1

Answer: (23.53, 25.47)

Step-by-step explanation:

The confidence interval for population mean(when population standard deviation is unknown) is given by :-

$\overline{x} \pm\ t^*\dfrac{s}{\sqrt{n}}$

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

n = sample size.

s = sample standard deviation.

t* = Critical value.

Given : $\overline{x}=24.5$

s= 4.4

Confidence level = 90% =0.090

Significance level = $\alpha=1-0.90=0.10$

Sample size : n= 58

Degree of freedom : df= n-1= 57

Using t-distribution table , the critical value :

$t_{\alpha/2,\ df}= t_{0.05,\ 57}=1.6720$

Then, the confidence interval will be :-

$24.5 \pm\ (1.6720)\dfrac{4.4}{\sqrt{58}}$

$24.5 \pm\ (1.6720)\dfrac{4.4}{7.61577310586}$

$24.5 \pm\ 0.965995165263\approx24.5\pm0.97\\\\=(24.5-0.97,\ 24.5+0.97)\\\\=(23.53,\ 25.47)$

Hence, a 90% confidence interval for the population mean = (23.53, 25.47)