\"What do you think is the ideal number of children for a family to have?\" A Ga
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Question
"What do you think is the ideal number of children for a family to have?" A Gallup Poll asked this question of 1016 randomly chosen adults. Almost half (49%) thought two children was ideal.† We are supposing that the proportion of all adults who think that two children is ideal is p = 0.49. What is the probability that a sample proportion p falls between 0.46 and 0.52 (that is, within ±3 percentage points of the true p) if the sample is an SRS of size n = 450? (Round your answer to four decimal places.) What is the probability that a sample proportion p falls between 0.46 and 0.52 if the sample is an SRS of size n = 5000? (Round your answer to four decimal places.) Combine these results to make a general statement about the effect of larger samples in a sample survey. Larger samples give a smaller probability that p will be close to the true proportion p. Larger samples have no effect on the probability that p will be close to the true proportion p. Larger samples give a larger probability that p will be close to the true proportion p.
Explanation / Answer
A)
Here,
n = 450
p = 0.49
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 0.46
x2 = upper bound = 0.52
u = mean = p = 0.49
s = standard deviation = sqrt(p(1-p)/n) = 0.023565512
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -1.273046841
z2 = upper z score = (x2 - u) / s = 1.273046841
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.10150071
P(z < z2) = 0.89849929
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.796998579 [ANSWER]
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b)
Here,
n = 5000
p = 0.49
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 0.46
x2 = upper bound = 0.52
u = mean = p = 0.49
s = standard deviation = sqrt(p(1-p)/n) = 0.007069653
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -4.24348947
z2 = upper z score = (x2 - u) / s = 4.24348947
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 1.10035E-05
P(z < z2) = 0.999988996
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.999977993 [ANSWER]
********************
c)
As we can see, it is
OPTION C: Larger samples give a larger probability that p will be close to the true proportion p. [ANSWER]
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