7. n = 256 8. A local board of education conducted a survey of residents in the

Question

7.

n = 256

8.

A local board of education conducted a survey of residents in the community concerning a property tax levy on the coming local ballot. They randomly selected 850 residents in the community and contacted them by telephone. Of the 850 residents surveyed, 410 supported the property tax levy. Let p represent the proportion of residents in the community that support the property tax levy. How large a sample n would you need to estimate p with margin of error 0.04 with 95% confidence? Assume that you don't know anything about the value of p. n = 601 n = 1037 n = 423

n = 256

8.

According to the National Institute on Alcohol Abuse and Alcoholism, and the National Institutes of Health, 41% of college students nationwide engage in binge drinking behavior, having five or more drinks on one occasion during the past two weeks. A college president wonders if the proportion of students enrolled at her college that binge drink is lower than the national proportion. In a commissioned study, 462 students are selected randomly from a list of all students enrolled at the college. Of these, 162 admitted to having engaged in binge drinking. The college president is more interested in testing her suspicion that the proportion of students at her college that binge drink is lower than the national proportion of 0.41. Her staff tests the hypotheses H0: p = 0.41, Ha: p < 0.41. The P-value is: between 0.025 and 0.05. between 0.01 and 0.025. between 0.05 and 0.10. below 0.01

Explanation / Answer

Question-7

Sample size n>= p*(1-p)*(z/ME)2=0.5*(1-0.5)*(1.96/0.04)2= 600.25

Hence Sample size n=601

Question-8

From following results the P-value is

Below 0.01

Test and CI for One Proportion

Test of p = 0.41 vs p < 0.41

Sample    X    N Sample p 95% Upper Bound Z-Value P-Value

1       162 462 0.350649         0.387165    -2.59    0.005

Using the normal approximation.

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