Homework SourseSuppose candidate A is favored over candidate B by 52 to 48
Suppose candidate A is favored over candidate B by 52% to 48%. A random sample of voters is selected. Use Excel and PHstat.
a. If the number in the sample is 50, what is the probability that the sample will indicate that candidate B wins(erroneously)?
b. If the number in the sample is 100, what is the probability that the sample will indicate that candidate B wins(erroneously)?
c. Determine the minimum sample size needed to insure the probability of an erroneous result would be less than 5%?
d. Determine the minimum sample size needed to insure the probability of an erroneous result would be less than 2%?
Solution
a)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 0.5
u = mean = n p = 0.48
s = standard deviation = sqrt(p(1-p)/n) = 0.070654087
Thus,
z = (x - u) / s = 0.283069259
Thus, using a table/technology, the right tailed area of this is
P(z > 0.283069259 ) = 0.388561873 [ANSWER]
*******************
b)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 0.5
u = mean = n p = 0.48
s = standard deviation = sqrt(p(1-p)/n) = 0.049959984
Thus,
z = (x - u) / s = 0.400320385
Thus, using a table/technology, the right tailed area of this is
P(z > 0.400320385 ) = 0.344460278 [ANSWER]
********************
c)
The z score corresponding to a right tailed area of 0.05 is
z = 1.644853627
Thus, the standard deviation is
s = (x- u)/z = (0.50-0.48)/1.644853627 = 0.012159137
Hence,
0.012159137 = sqrt(p(1-p)/n)
0.012159137 = sqrt(0.48*(1-0.48)/n)
n = 1688.259115
Rounding up,
n = 1689. [ANSWER]
************************
d)
The z score corresponding to a right tailed area of 0.05 is
z = 2.053748911
Thus, the standard deviation is
s = (x- u)/z = (0.50-0.48)/2.053748911 = 0.009738289
Hence,
0.009738289 = sqrt(p(1-p)/n)
0.009738289 = sqrt(0.48*(1-0.48)/n)
n = 2631.959984
Rounding up,
n = 2632 [ANSWER]
