Question

The magazine Sports Illustrated asked a random sample of 750 Division I college athletes, "Do you believe performance-enhancing drugs are a problem in college sports?" Suppose that 30% of all Division I athletes think that these drugs are a problem. Let p^ be the sample proportion who say that these drugs are a problem.

Which of the following are the mean and standard deviation of the sampling distribution of the sample proportion p^?

a) Mean = 0.30, SD = 0.017
b) Mean = 0.30, SD = 0.55
c) Mean = 0.30, SD = 0.0003
d) Mean = 225, SD = 12.5
e) Mean = 225, SD = 157.5

Answer 1

The mean of the sampling distribution of the sample proportion is 0.30, and the standard deviation is found to be approximately 0.017. Therefore, the correct answer is Mean = 0.30, SD = 0.017.

Step-by-step explanation:

To determine the mean and standard deviation of the sampling distribution of the sample proportion p, we use the formulas related to binomial distributions, since the survey outcome (believe that performance-enhancing drugs are a problem or not) follows a binomial distribution. The mean of the sampling distribution of the proportion is simply the population proportion, which is given as 0.30.

The standard deviation (SD) of the sampling distribution of p can be calculated using the formula SD = √[p(1-p)/n], where p is the population proportion and n is the sample size. In this case:

SD = √[0.30(1-0.30)/750] = √[0.21/750] = √[0.00028] ≈ 0.0167

Thus, the correct answer is a) Mean = 0.30, SD = 0.017

Answer 2

a) Mean = 0.30, SD = 0.017

Step-by-step explanation:

The mean and sampling distribution of the sample proportion can be found using the equations

Mean= p

`SD=\sqrt{(p*(1-p))/(n) }`

where

• n is the sample size (750)

• p is the sample proportion of all Division I athletes think that these drugs are a problem (0.30 or 30%)

Using these information:

Mean = 0.30

`SD=\sqrt{(0.30*0.70)/(750) }` ≈ 0.017