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. Of the 5594Division Imale athletes who responded to the survey, 3547reported 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

estimate 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?

Short Answer

Expert verified

a. Confidence level is 87.88%

b. Non Response Bias.

Step by step solution

01

Given Information

It is given that n=5594

x=3547

E=1%=0.01

02

Calculation of Confidence Level

Sample Proportion is calculated as:

p^=xn

p^=35475594=0.6341

Margin of error is calculated as:

E=Zα/2×p^(1-p^)n

0.01=Zα/2×0.6341(1-0.6341)5594

Za/2=0.010.00644=1.55

Hence, Probability:

P(-1.55<Z<1.55)=P(Z<1.55)-P(Z<-1.55)

P(-1.55<Z<1.55)=0.9394-0.0606

P(-1.55<Z<1.55)=0.8788=87.88%

Hence, confidence level is87.88%

03

Bias in Study

Non Sampling Bias are:

Selection Bias, Response Bias, Non Response Bias.

If we don't have data for everybody in sample, it is result of non response bias.

So, we cannot generalize the result as we don't have sufficient data.

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Most popular questions from this chapter

The U.S. Forest Service is considering additional restrictions on the number of vehicles allowed to enter Yellowstone National Park. To assess public reaction, the service asks a random sample of 150 visitors if they favor the proposal, Of these, 89 say "Yes."

a. Construct and interpret a 99%confidence interval for the proportion of all visitors to Yellowstone who favor the restrictions.

b. Based on your work in part (a), is there convincing evidence that more than half of all visitors to Yellowstone National Park favor the proposal? Justify your answer.

A quality control inspector will measure the salt content (in milligrams) in a random

sample of bags of potato chips from an hour of production. Which of the following would result in the smallest margin of error in estimating the mean salt content μ?

a. 90%confidence; n=25

b. 90%confidence; n=50

c. 95%confidence; n=25

d. 95%confidence; n=50

e. n = 100 at any confidence level

School vouchers A small pilot study estimated that 44%of all American

adults agree that parents should be given vouchers that are good for education at any public or private school of their choice.

a. How large a random sample is required to obtain a margin of error of at most 0.03with 99%confidence? Answer this question using the pilot study’s result as the guessed value for p^

b. Answer the question in part (a) again, but this time use the conservative guessp^=0.5. By how much do the two sample sizes differ?

In a poll conducted by phone,

I. Some people refused to answer questions.

II. People without telephones could not be in the sample.

III. Some people never answered the phone in several calls.

Which of these possible sources of bias is included in the ±2%margin of error announced for the poll?

a. I only

b. II only

c. III only

d. I, II, and III

e. None of these

How confident? The figure shows the result of taking 25SRSs from a Normal population and constructing a confidence interval for the population mean using each sample. Which confidence level—80%,90%,95%,or99%—do you think was used? Explain your reasoning.

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