Question

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%?

Answer #1

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 = 0.01321^{2}

n = 0.2496 / 0.01321^{2}

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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