Suppose candidate A is favored over candidate b by 52% to 48%. A random sample of voters is selected.
A)if the number in the sample is 70, what is the probability that the sample will indicate that candidate b wins (erroneously)?
B)Determine the minimum sample size needed to insure the probability of an erroneous results would be less than 6.5%?
A)
Let p be the proportion of voters favored candidate A.
Then point estimate of p is 0.52
Standard error of the proportion, s =
Probability that the sample will indicate that candidate b wins = P(p < 0.50)
= P[Z < (0.5 - 0.52) / 0.0597]
= P[Z < -0.335]
= 0.3688
B)
Let n be the sample size and s be the standard error of the proportion.
Probability of an erroneous results would be less than 6.5%
P(p < 0.50) = 0.065
P[Z < (0.50 - 0.52)/s] = 0.065
(0.50 - 0.52)/s = -1.514
s = -0.02 / -1.514
s = 0.01321
0.2496 / n = 0.013212
n = 0.2496 / 0.013212
n = 1430.339 1431 (Rounding to next integer)
The minimum sample size needed to insure the probability of an erroneous results would be less than 6.5% is 1431.
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