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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