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)

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(s)/(√(n))`

`19.1\pm (3.355)(1.5)/(√( 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) .