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Suppose public opinion is split 65% against and 35% for increasing taxes to help

ID: 3131707 • Letter: S

Question

Suppose public opinion is split 65% against and 35% for increasing taxes to help balance the federal budget. Suppose also that 500 people from the population are selected randomly and interviewed.

(a)[3] Completely describe the sampling distribution of the sample proportion of people who are in favour of increasing taxes to help balance the federal budget.

(b)[3] What is the probability the proportion favouring a tax increase is more than 30%? (c) [4]Construct a 95% confidence interval for proportion of people who are in favor of

increasing taxes to help balance the federal budget

Explanation / Answer

a)

Here,  
n =    500
p =    0.35

Hence, the sample proportions are approximately normally distributed with mean

u = mean = p =    0.35

and standard deviation of
  
s = standard deviation = sqrt(p(1-p)/n) =    0.021330729. [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.3      
u = mean = p =    0.35      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.021330729      
          
Thus,          
          
z = (x - u) / s =    -2.344036155      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -2.344036155   ) =    0.990461839 [ANSWER]

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c)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.35          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.021330729          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
Margin of error = z(alpha/2)*sp =    0.041807461          
lower bound = p^ - z(alpha/2) * sp =   0.308192539          
upper bound = p^ + z(alpha/2) * sp =    0.391807461          
              
Thus, the confidence interval is              
              
(   0.308192539   ,   0.391807461   ) [ANSWER]

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