Question

A random sample of 144 observations has a mean of 20, a median of 21, and a mode of 22. The population standard deviation is known to equal 4.8. The 95.44% confidence interval for the population mean is______?
a. 19.200 to 20.800
b. 18.78 to 20.784
c. 15.2 to 24.8
d. 19.216 to 20.784
e. 21.2 to 22.8

Answer 1

Answer: a. 19.200 to 20.800

Step-by-step explanation:

Formula to find the confidence interval$(\mu)$ :-

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

, where n is the sample size

$\sigma$ = Population standard deviation.

$\overline{x}$= Sample mean

$z_{\alpha/2}$ = Two tailed z-value for significance level of $\alpha$.

Given : Confidence level = 95.44% = 0.9544

Significance level = $\alpha=1-0.9544=0.0456$

Using standard normal z-value table ,

Two tailed z-value for Significance level of 0.0456 :

$z_{\alpha/2}=z_{0.0228}=1.999\approx2$

Also,

n=144

$\sigma= 4.8$

$\overline{x}=20$

Then, the required 95.44% confidence interval for the population mean :-

$20\pm (2)\dfrac{4.8}{\sqrt{144}}\\\\ =20\pm (0.800)\\\\=(20-0.800,\ 20+0.800)=(19.200,\ 20.800)$

Hence, the 95.44% confidence interval for the population mean is 19.200 to 20.800.