Question

A student wanted to construct a 95% confidence interval for the average age of students in her statistics class. She randomly selected 9 students. Their average age was 19.1 years with a standard deviation of 1.5 years. What is the 99% confidence interval for the population mean?

Answer 1

Answer: (17.42, 20.78)

Step-by-step explanation:

As per given , we have

Sample size : n= 9

$\overline{x}=19.1$ years

Population standard deviation is not given , so it follows t-distribution.

Sample standard deviation : s= 1.5 years

Confidence level : 99% or 0.99

Significance level : $\alpha= 1-0.99=0.01$

Degree of freedom : df= 8  (∵df =n-1)

Critical value : $t_c=t_{(\alpha/2,\ df)}=t_{0.005,\ 8}= 3.355$

The 99% confidence interval for the population mean would be :-

$\overline{x}\pm t_c\dfrac{s}{\sqrt{n}}$

$19.1\pm (3.355)\dfrac{1.5}{\sqrt{ 9}}\\\\=19.1\pm1.6775\\\\=(19.1-1.6775,\ 19.1+1.6775)\\\\=(17.4225,\ 20.7775)\approx(17.42,\ 20.78)$

Hence, the 99% confidence interval for the population mean is (17.42, 20.78) .