BioStats: Sampling Distributions Help! My professor did not cover this in class

Question

BioStats: Sampling Distributions Help!

My professor did not cover this in class and our homework is due tonight. I really am not understanding the concept and would appreciate any guidance on how to think about this and what equations would be relevant. Thank you very much in advance!

"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.

(a) 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 = 200? (Round your answer to four decimal places.)

(b) 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.)

(c) Combine these results to make a general statement about the effect of larger samples in a sample survey.

__ Larger samples have no effect on the probability that p will be close to the true proportion p.

__Larger samples give a smaller 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 =    200      
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) =    0.035348267      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.848697894      
z2 = upper z score = (x2 - u) / s =    0.848697894      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.198024709      
P(z < z2) =    0.801975291      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.603950582   [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]

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c)

As we can see,

OPTION C: Larger samples give a larger probability that p will be close to the true proportion p. [ANSWER]

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d)

As

sigma(p^) = sqrt(p(1-p)/n)

Then

n = p(1-p)/sigma^2 = 0.44*(1-0.44)/0.004^2 = 15400 [ANSWER]

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