A simple random sample of 150 university students found that 40 of them reported

Question

A simple random sample of 150 university students found that 40 of them reported smoking tobacco.

a) Verify that the three conditions for normal approximation are fufilled, and estimate the proportion of all students that smoke.

b) What type of bias, response or non-response, is most likely to affect your estimate of the smoking proportion? Is this bias likely to make your estimate lower or higher than the true value?

c) The university intends to start an anti-smoking campaign if more than 20% of the students smoke. Using the alpha = 0.05 significance level, is there sufficient enough evidence to start such a campaign?

d) Use the provided R-code of the p-value from the binomial distribution. Is the normal approximation (the p- value you found in part c) reasonably close?

e) Imagine that we had only taken an SRS of 15 people and that 4 of those had reported smoking. Would the three conditions hold?

f) Use the normal approximation to find the p-value against the null that 20% or less of the student population smokes, like you did in part c. Use the provided R-code to find the p-value from the binomial distribution. How close are these values this time?

Explanation / Answer

n = 150 , X = 40

n p = X = 40 > 10

nq = n -X = 110 > 10

hence

2) non-response bias   is most likely to affect your estimate of the smoking proportion .

this bias is likely to make estimate lower than the true value

because those who smoke amy not like to response

3) p^ = X/n = 40/150 = 0.26666

Ts = (p^ - 0.20)/sqrt(0.2*0.8/150) = (0.2666-0.2)/sqrt(0.2*0.8/150) =2.0392

p-value = P(Z > 2.0392) = 0.0207

d) p-value from binomial = 0.0186968

yes ,result are closer

e) np =X = 4 < 10

hence conditions don't hold

f) this time ,results are not closer

from normal - P(Z> (4/15 - 0.2)/(sqrt(0.2*0.8/15)) = P(Z> 0.644851) = 0.2595

from binomial - 0.164233

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