Question

A random sample of 125 students is chosen from a population of 4,000 students. If the mean IQ in the sample is 100 with a standard deviation of 8, what is the 98% confidence interval for the students' mean IQ score?

Answer 1

Hiya! The answer is 98.33−101.67 :)

Answer 2

Answer: 98.33 - 101.67

Explanation:

The given Sample mean n= 100

Standard deviation
`\sigma=8`

Mean IQ in the sample
`\bar{x}=125`

Standard error of mean =
`\frac{\sigma}{\sqrt{\bar{x}}}`

Standard error of mean =
`(8)/(√(125))`

Thus, standard error =
`(8)/(11.180)`

Standard error of mean (SE)= 0.7155

We know that z - score for 98% confidence interval is 2.33 .

Now, 98% confidence interval will be

n-z(SE) and n+z(SE)

`=100-(0.7155)(2.33)\ and\ 100+(0.7155)(2.33)`

`=100-1.66726\ and\ 100+1.66726`

`=98.33274\ and\ 101.667`

Hence, the 98% confidence interval for the students' mean IQ score will be 98.33 - 101.67