Students conducted a survey and found out that 36% of their peers on campus had
ID: 3223879 • Letter: S
Question
Students conducted a survey and found out that 36% of their peers on campus had tattoos, but only 4% of their peers were smokers. If 200 students were surveyed, can these students use the Normal approximation to study the proportion of students in the population who have tattoos? no, because either np or n(1 - p) are less than 15 yes, because both np and n(1 - p) are less than 15 no, because either np or n(1 - p) are greater than 15 yes, because both np and n(1 - p) are greater than 15 Students conducted a survey and found out that 36% of their peers on campus had tattoos, but only 4% their peers were smokers. The standard error or hat p of tattooed students in the 100-student sample is: 5%, 4%, 6%, It is not possible to tell from the information provided. Each person in a random sample of 2000 "likely voters" (as defined by a professional polling organization) was questioned about his or her political views. Of those surveyed, 1308 felt that "the economy's state" was the most urgent national concern. A 99% confidence interval for the proportion p of all likely voters that feel the economy's state is the most urgent national concern is given by (use the plus four confidence interval procedure): 0.624 to 0.663. 0.627 to 0.681. 0.615 to 0.672. 0.606 to 0.680. 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 A 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 = 256 n = 1037 n = 423 n = 601 A veterinarian studying causes of in houses suspects feeding too much be a culprit. The veterinarian hypothesizes that more than two thirds of all horses with enter which are fed mostly alfalfa. The proper hypothesis statement is: H_0: p = 0.67, H_a: p > 0.67. H_0: p = 0.67, H_a: pExplanation / Answer
a.
Here n=200and p=0.36
n*p=200*0.36=72
n*p*q=200*0.36*(1-0.36)=46.08
Both np, npqare greater than 15. We can use approximale proportion of student in the population who have tattoos by using Normal approximation.
4th option is correct. Yes np and np(1-p) are greater than 15
b.
the standard error can be approximated by:
SEp = sqrt[ p * ( 1 - p ) / n ]
=sqrt[0.36*(1-0.36)/100]
=0.048~5%
First option is correct.
3.
Using Minitab:
Test and CI for One Proportion
Sample X N Sample p 99% CI
1 1308 2000 0.654000 (0.626006, 0.681253)
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