Question

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

Answer 1

Answer:
`(18.17,\ 20.03)`

Explanation:

Given : Sample size :
`n=9` , which is a small sample (<30), so we use t-test.

Sample mean :
`\overline{x}=19.1\text{ years}`

Standard deviation :
`\sigma =1.5\text{ years}`

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

Critical value :
`t_(n-1, \alpha/2)=t_(8, 0.005)=1.86`

The formula to find the confidence interval for population mean :-

`\overline{x}\pm t_(n-1,\alpha/2)(\sigma)/(√(n))\\\\=19.1\pm(1.86)(1.5)/(√(9))\\\\\approx19.1\pm0.93\\\\=(19.1-0.93,\ 19.1+0.93)\\\\=(18.17,\ 20.03)`

Hence,the 99% confidence interval for the population mean =
`(18.17,\ 20.03)`