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 int…

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asked Mar 15, 2018 118k views

2 Answers

Boyan Hristov · Answer 1 · 2018-03-20T22:56:42+0000

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