Question

A study was recently conducted to estimate the mean cholesterol for adult males over the age of 55 years. From a random sample of n10 ​men, the sample mean was found to be 242.6 and the sample standard deviation was 73.33. To find the 95 percent confidence interval estimate for the​ mean, what is the correct critical value to​ use?

Answer 1

Answer:  Correct critical value = 2.2622

Step-by-step explanation:

Confidence interval for population mean when population standard deviation is unknown:

$\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}$ , where $\overline{x}$ = sample mean, n= sample size, s= sample standard deviation, $t_{\alpha/2}$ = two-tailed t value.

As per given: n= 10

degree of freedom : df = n-1=9

$\overline{x}=242.6\\\\s=73.33\\\\\alpha=5\%=0.05$

Critical t-value : $t_{\alpha/2,df}=t_{0.05/2,9}=t_{0.025,9}=2.2622$

So, the 95 percent confidence interval estimate for the​ mean :

$242.6\pm (2.2622)\dfrac{73.33}{\sqrt{10}}\\\\=242.6\pm (2.2622)(23.19)\\\\\approx242.6\pm52.46\\\\=(242.6-52.46,\ 242.6+52.46)\\\\=(190.14,\ 295.06)$

The 95 percent confidence interval estimate for the​ mean:(190.14, 295.06)