Question

There were 800 math instructors at a mathematics convention. Forty instructors were randomly selected and given an IQ test. The scores produced a mean of 130 with a standard deviation of 10. Find a 95% confidence interval for the mean of the 800 instructors. Use the finite population correction factor.

Answer 1

Answer: The required interval would be (127.711, 132.289).

Explanation:

Since we have given that

N = 800

n = 40

Mean = 130

Standard deviation = 10

degree of freedom df = n-1 = 40 -1 = 39

At 95% confidence level,
`t_(39)=2.023`

So, 95% confidence interval would be

`\bar{x}\pm t_(n-1)(s)/(√(n))\sqrt{(N-1)/(n-1)}\\\\=130\pm 2.023* \frac{10}{\sqt{40}}\sqrt{(800-1)/(40-1)}\\\\=130\pm 2.023* (1)/(4)\sqrt{(799)/(39)}\\\\=130\pm 2.289\\\\=(130-2.289,130+2.289)\\\\=(127.711,132.289)`

Hence, the required interval would be (127.711, 132.289).

Answer 2

95% confidence interval for the mean of the 800 instructors is [126.80 , 133.20].

Explanation:

We are given that there were 800 math instructors at a mathematics convention.

Forty instructors were randomly selected and given an IQ test. The scores produced a mean of 130 with a standard deviation of 10.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. =
`(\bar X-\mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where,
`\bar X` = sample mean score = 130

s = sample standard deviation = 10

n = sample of instructors = 40

`\mu` = population mean of 800 instructors

Here for constructing 95% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.

So, 95% confidence interval for the population mean,
`\mu` is ;

P(-2.0225 <
`t_3_9` < 2.0225) = 0.95 {As the critical value of t at 39 degree of

freedom are -2.0225 & 2.0225 with P = 2.5%}

P(-2.0225 <
`(\bar X-\mu)/((s)/(√(n) ) )` < 2.0225) = 0.95

P(
`-2.0225 * {(s)/(√(n) ) }` <
`{\bar X-\mu}` <
`2.0225 * {(s)/(√(n) ) }` ) = 0.95

P(
`\bar X-2.0225 * {(s)/(√(n) ) }` <
`\mu` <
`\bar X+2.0225 * {(s)/(√(n) ) }` ) = 0.95

95% confidence interval for
`\mu` = [
`\bar X-2.0225 * {(s)/(√(n) ) }` ,
`\bar X+2.0225 * {(s)/(√(n) ) }` ]

= [
`130-2.0225 * {(10)/(√(40) ) }` ,
`130+2.0225 * {(10)/(√(40) ) }` ]

= [126.80 , 133.20]

Therefore, 95% confidence interval for the mean of the 800 instructors is [126.80 , 133.20].