A political pollster is conducting an analysis of sample results in order to mak
ID: 3221896 • Letter: A
Question
A political pollster is conducting an analysis of sample results in order to make predictions on election night. Assuming a two-candidate election, if a specific candidate receives et least 53% of the vote in the sample, that candidate will be forecast as the winner of the election. You select a random sample of 100 voters. Complete parts (a) through (c) below. c. What is the probability that a candidate will be forecast as the winner when the population percentage of her vote is 49% (and she will actually lose the election)? The probability is that a candidate will be forecast as the winner when the population percentage of her vote is 49%. (Round to tour decimal places as needed.) Suppose that the sample size was increased to 400. Repeat process (a) through (C), using this new sample size. Comment on the difference. The probability is that a candidate will be forecast as the winner when the population percentage of her vote is 50.1%. (Round to four decimal places as needed.) The probability is that a candidate will be forecast as the winner when the population percentage of her vote is 50%. (Round to tour decimal places as needed.) The probability is at a candidate will be forecast as the winner when the population percentage of her vote is 49%. Choose the correct answer below. A. Increasing the sample size by a factor of 4 decreases the standard error by a factor 2. Changing the standard error doubles the magnitude of the standardize Z-value. B. increasing the sample size by a factor of 4 increases the standard error by a factor of 2. Changing the standard error decreases the standardized Z-value to half or its original value. C. Increasing the sample size by a factor of 4 decreases the standard error by a factor of 2. Changing the standard error decreases the standardized Z-value to half of its original value. D. Increasing the sample size by a factor of 4 increases the standard error by a factor of 2. Changing the standard error doubles the magnitude of the standardized Z-value.Explanation / Answer
c) n = 100 , p^ = 0.49
P(p>0.53)
Z =(p -p^)/(sqrt(p^*q^/n)
P (p> 0.53) = P( z> ((0.53-0.49)/sqrt(0.49*0.51/100))) = P (Z > 0.80) = 0.2119
d)i) p^ = 0.501 , n= 400
P (p> 0.53) = P( z> ((0.53-0.501)/sqrt(0.499*0.501/400))) = P (Z > 1.16) = 0.123
ii)p^ = 0.56 n = 400
P (p> 0.53) = P( z> ((0.53-0.56)/sqrt(0.56*0.44/400))) = P (Z > -1.208) = 0.8865
iii)p^ = 0.49 ,n = 400
P (p> 0.53) = P( z> ((0.53-0.49)/sqrt(0.49*0.51/400))) = P (Z > 1.60) =0.0548
e)Increasing the sample size by a factor of 4 decreases the standard error by a factor of 2 and double the standard z-value
hence option a) is correct