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:

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

Step-by-step 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. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

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 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.0225) = 0.95

P( $-2.0225 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.0225 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.0225 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.0225 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.0225 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.0225 \times {\frac{s}{\sqrt{n} } }$ ]

                   = [ $130-2.0225 \times {\frac{10}{\sqrt{40} } }$ , $130+2.0225 \times {\frac{10}{\sqrt{40} } }$ ]

                   = [126.80 , 133.20]

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

Answer 2

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

Step-by-step 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

$\bar{x}\pm t_{n-1}\dfrac{s}{\sqrt{n}}\sqrt{\dfrac{N-1}{n-1}}\\\\=130\pm 2.023\times \dfrac{10}{\sqt{40}}\sqrt{\dfrac{800-1}{40-1}}\\\\=130\pm 2.023\times \dfrac{1}{4}\sqrt{\dfrac{799}{39}}\\\\=130\pm 2.289\\\\=(130-2.289,130+2.289)\\\\=(127.711,132.289)$

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