Q.48 - The Practice of Statistics for AP

Question

Gambling and the NCAA Gambling is an issue of great concern to those involved in college athletics. Because of this concern, the National Collegiate Athletic Association (NCAA) surveyed randomly selected student-athletes concerning their gambling-related behaviors.$17$ Of the $5594$ Division I male athletes in the survey, $3547$ reported participation in some gambling behavior. This includes playing cards, betting on games of skill, buying lottery tickets, betting on sports, and similar activities. A report of this study cited a $1$% margin of error.

(a)  The confidence level was not stated in the report. Use what you have learned to find the confidence level, assuming that the NCAA took an SRS.

(b)  The study was designed to protect the anonymity of the student-athletes who responded. As a result, it was not possible to calculate the number of students who were asked to respond but did not. How does this fact affect the way that you interpret the results?

Step-by-Step Solution

Answer

From the given information,

a) The sample size of $318$ has tested PTC when the sample proportion $\stackrel{\wedge }{p} is 75\%$

b) The sample size is increased by $105$ when the guessed value $\stackrel{\wedge }{p} is 0.5$

Part (a) Step 1: Given Information

It is given in the question that, 

The margin of error $E=0.04$

Guessed value $\stackrel{\wedge }{p} =75\%$

Confidence interval $=90\%$

The confidence level was not stated in the report. Use what you have learned to find the confidence level, assuming that the NCAA took an SRS. 

Part (a) Step 2: Explanation

The confidence interval is $90\%$.

Convert $90\%$into a decimal.

$\frac{90}{100}=0.90$

For confidence interval $0.90$, use table A.

$z_{\alpha /2}=1.645$

Guessed value $\stackrel{\wedge }{p}=75\%$

convert $75\%$ into decimal.

$\frac{75}{100}=0.75$

Now, find the sample size. Use the formula $n=\frac{{\left[z_{\alpha /2}\right]}^2\hat{p}(1-\hat{p})}{E^2}$

$n=\frac{{\left[z_{\alpha /2}\right]}^2\hat{p}(1-\hat{p})}{E^2}$

$n=\frac{{1.645}^2\times 0.75\times (1-0.75)}{{0.04}^2}$

$=318$

Therefore the sample size is $318$

Part (b) Step 1: Given Information

it is given in the question that,

The margin of error $E=0.04$

Sample proportion $\stackrel{\wedge }{p}=0.5$

Confidence interval $=90\%$

It was not possible to calculate the number of students who were asked to respond but did not. How does this fact affect the way that you interpret the results?

Part (b) Step 2: Explanation

The confidence interval is $90\%$.

Convert $90$% into a decimal.

$\frac{90}{100}=0.90$

For confidence interval $0.90$, use table A.

$z_{\alpha /2}=1.645$

From part(a) sample size $n=318$

Find the new sample size. Use the formula $n=\frac{{\left[z_{\alpha /2}\right]}^2\hat{p}(1-\hat{p})}{E^2}$

$n\text{'}=\frac{{\left[z_{\alpha /2}\right]}^2\hat{p}(1-\hat{p})}{E^2}$

$n^{\mathrm{\prime }}=\frac{{1.645}^2\times 0.5\times (1-0.5)}{{0.04}^2}$

$=423$

Therefore, the new sample size is $423$.

For the difference between the sample sizes, subtract $n$ from $n\text{'}$

$n^{\mathrm{\prime }}-n=423-318$

$=105$

Hence, the sample size is increased by $105$ .