Question

Assuming the population of interest is approximately normally​ distributed, construct a 95​% confidence interval estimate for the population mean given the values below. x overbar=19.3 s=3.1 n=23 The 95​% confidence interval for the population mean is from nothing to nothing.

Answer 1

Answer: $(17.96,20.64)$

Step-by-step explanation:

The confidence interval for population mean is given by :-

$\overline{x}\ \pm\ t_{n-1,\alpha/2}\dfrac{s}{\sqrt{n}}$

Given : $\overline{x}=19.3$

$s= 3.1$

n=23, which is a small sample(n<30), so we use t-test.

Significance level: $1-0.95=0.05$

Critical value : $t_{n-1,\alpha/2}=t_{22,0.025}=2.074$

Then , the  confidence interval for population mean will be  :-

$19.3\ \pm\ (2.074)\dfrac{3.1}{\sqrt{23}}\\\\\approx19.3\pm1.34\\\\=(19.3-1.34,19.3+1.34)\\\\=(17.96,20.64)$

Hence, the 95​% confidence interval for the population mean is $(17.96,20.64)$