Question

ch 22 tv news program conducts a call - in poll about a proposed city ban on smoking in public places. of the 2467 callers, 1900 were opposed to the ban. which of the following statements are true with respect to using this sample to estimate p, the proportion of all tv news viewers that favor such a ban on smoking in public places?
there appear to be no violations to any of the assumptions made in using the methods of this chapter. we should now be able to use this sample to estimate p.
n is so large that both the count of successes (np^), and the count of failures(n(1 - p^)), are 15 or more, so its okay to use this sample to estimate p.
the population is much larger than the sample, so its okay to use this sample to estimate p.
there is no way this sample can be viewed as an srs of all tv news viewers, so we cant use this sample to estimate p.
question 10
1 pts
ch 22 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 100 students were surveyed, can these students use the normal approximation to study the proportion of students in the population who have tattoos? (note: np is the number of successes, n(1 - p) is the number of failures.)
yes, because both np and n(1 - p) are less than 15.
no, because either np or 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.

Explanation:

Step1: Calculate \( np \) and \( n(1 - p) \)

Given \( n = 100 \) and \( p=0.36 \) (proportion of students with tattoos).
\( np=100\times0.36 = 36\)
\( n(1 - p)=100\times(1 - 0.36)=100\times0.64 = 64\)

Step2: Check the condition for normal approximation

For normal approximation to the binomial distribution (to study the proportion \( p \)), the conditions are \( np\geq15 \) and \( n(1 - p)\geq15 \).
Since \( np = 36\geq15 \) and \( n(1 - p)=64\geq15 \)

Answer:

Yes, because both \( np \) and \( n(1 - p) \) are greater than \( 15 \).