Question

"A public relations officer of William Paterson University wants to estimate the mean IQ of the university students. If she wants to be 99% confident that her sample mean is off by no more than 3 points, how many students she has to test in order to come up with a valid estimation? A recent study shows that IQ of New Jersey students has standard deviation of 15 points a) 162 b) 13 c) 171 d) None of the above A researcher wants to determine the proportion of pro-peace students at Rutgers University. He has no idea what the sample proportion will be. How large a sample is required to be 95% sure that the sample proportion is off by no more than 4%? a) 68 b) 849 c) 595 d) 601 e) None of the above *The marks on a statistics test are normally distributed with a mean of 62 and a variance of 225. If the instructor wishes to assign Bs or higher to the top 20% of the students in the class, what mark is required to get a B or higher? a) 68.7 b) 71.5 c) 74.6 d) 49.4 e) 69.9 *A company has a machine that produces cans of coconut milk, and it has been noticed that the amount of coconut milk varies from can to can. The amounts are normally distributed with standard deviation 7 milliliters. The label used on the cans states that each can contains 412 milliliters. The management of the company decides to set the mean μ of the amount of coconut milk per can so that 80% of the cans contain more than 412 milliliters. Of the following, which is the closest approximation to μ in milliliters? a) 413 b)415 c) 418 d) 403 e) 406 f) None of the above

Answer 1

1)

Confidence level = 0.99

Zc = 2.58 ( Using z table)

Margin of error = E = 3

Population standard deviation = $\sigma$ = 15

We have to find sample size (n)

$n=\left ( \frac{Zc*\sigma}{E} \right )^{2}$

$n=\left ( \frac{2.58*15}{3} \right )^{2}$

$n=\left12.9^{2}=166$

2)

If we have not given a prior estimate we take p = 0.5 for unbiasedness.

Margin of error = E = 0.04

Confidence level = C = 0.95

Zc = 1.96 ( Using z table)

p = 0.50

We have to find sample size (n)

$n=p(1-p)\left ( \frac{Zc}{E} \right )^{2}$

$n=0.50(1-0.50)\left ( \frac{1.96}{0.04} \right )^{2} =0.25*49^{2}=600$

3)

We have given P(X > x) = 0.20

z value 0.20 is 0.84

We have to find value of x

$x=\mu+z*\sigma=62+0.84*15=74.6$

Therefore correct option is c) 74.6

4)

We have given P(X > x) = 0.80

z value for 80% is - 0.84

We have to find value of x

$x=\mu+z*\sigma=412-0.84*7=406$

Therefore correct option is e) 406