News companies often broadcast the result of polls saying things like \"56% of p
ID: 3263301 • Letter: N
Question
News companies often broadcast the result of polls saying things like "56% of people think pineapple is an acceptable topping for a pizza" with a plusminus 4% margin of error. This suggests that some sort of two-sided confidence interval was constructed. We plan to pole people uniformly at random1 asking them whether pineapple is an appropriate topping for pizza. Let X_1, ..., X_n be our random sample where X_i = 1 if the ith person approves. We now will give a two-sided 95% confidence interval for mu = p using our estimate p = X. (a) Is pineapple an acceptable topping for pizza? After all other questions are answered and time permitting, feel free to give an exposition on the merits (or lack of) of pineapple pizza. An exposition would not influence your grade. (b) Suppose we poll 60 Georgia Tech students uniformly at random and 40 of them say that pineapple is an appropriate topping for pizza. Give a 99% confidence interval for p. You can calculate it but the sample variance is sigma^2 = s^2 = sigma_i=1^50 (x_i - x)^2/59 = 20/59. (c) Using your answer to (b), do you feel comfortable saying "I'm 99% confident that the proportion of Georgia Tech students that find pineapple to be an acceptable topping is significantly different than 0.5"? Why or why not?Explanation / Answer
2)
b) X = 40 , n = 60
p^ = X/ n = 40/60 = 0.6667
z for 99 % is 2.576
hence confidence interval is
(2/3 - 2.576 * sqrt((20/59)/60) , (2/3 + 2.576 * sqrt(( ,20/59)/60)
=( 0.47304264, 0.86029)
c)since 0.5 is in the confidence interval
we don't feel comfortable saying 99 % confident that p is significant from 0.5