Last year, a national opinion poll found that 43% of all Americans agree that pa

Question

Last year, a national opinion poll found that 43% of all Americans agree that parents should be given vouchers good for education at any public or private school of their choice. Assume that in fact the population proportion is 0.43. This year, a random sample of 350 is to be selected and asked the same question. Please show your work.

A) What are the shape, mean(expected value), and standard deviation of the sampling distribution of the sample proportion for samples of 350?

B) What is the probability that the sample proportion will fall within 0.03 of the population proportion?

C) Redo part B), but assume that the sample size is 700. By how much does your answer change and WHY?

Explanation / Answer

a)

Here,          
n =    350      
p =    0.43      

Hence, it is approximately bell shaped, with

u = mean = p =    0.43      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.026462912   [ANSWER]

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b)

Here,          
n =    350      
p =    0.43      
We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound = 0.43-0.03 =   0.4      
x2 = upper bound = 0.43+0.03   0.46      
u = mean = p =    0.43      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.026462912      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.133662083      
z2 = upper z score = (x2 - u) / s =    1.133662083      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.12846816      
P(z < z2) =    0.87153184      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.74306368   [ANSWER]

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c)

Here,          
n =    700      
p =    0.43      
We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.4      
x2 = upper bound =    0.46      
u = mean = p =    0.43      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.018712105      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.603240293      
z2 = upper z score = (x2 - u) / s =    1.603240293      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.054440806      
P(z < z2) =    0.945559194      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.891118387   [ANSWER]

**********

Our answer increased by 0.891118387-0.74306368 = 0.148054707. It is because the sample size incrased, so there is less variation in p^, so it is more likely to get a proportion close to the population proportion.
  

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