Question

How do I determine a population deviation with literally only a percentage???

The question is phrased like this... Suppose that 44% of all Americans approve of the job the President is doing. The most recent Gallup poll had a sample size of 1400.

a) What is the mean of the sampling distribution
b) What is the standard deviation of the sampling distribution (Don't forget to justify the use of the formula).
c) Describe the normal approximation for this sampling distribution (Don't forget to justify this!) You can simply write as N ( mean, standard deviation).
d) What is the probability that the Gallup poll will come up with a probability within three percentage points of the true 44%?

When you answer, please give the answers AND a detailed explanation of how you got the answers.

Answer 1

Part A:

Given that 44% of all Americans approve of the job the President is doing. The most recent Gallup poll had a sample size of 1400.

Thus, p = 44% = 0.44 and n = 1400.

Mean of a sample distribution of proportions is given by

$\mu=np \\ \\ =1400\times0.44 \\ \\ =616$


Part B:

The standard deviation of a sample distribution of proportions is given by:

$\sigma=\sqrt{npq}=\sqrt{np(1-p)} \\ \\ =\sqrt{1400\times0.44\times(1-0.44)} \\ \\ =\sqrt{616\times0.56}=\sqrt{344.96}=18.57$


Part C:

The normal distribution is given by N(mean, standard deviation).
Given from part A, that the mean of the distribution is 616, and from part B, that the standard deviation of the distribution is 18.57. The normal approximation of the distribution is given by N(616, 18.57).

A normally distributed distribution with a mean of 616 and a standard deviation of 18.57.


Part D:

The probability that the Gallup poll will come up with a probability within three percentage points of the true 44% is given by:

$P(\hat{p}\ \textless \ 44\%\pm3\%)=P(\hat{p}\ \textless \ 0.44\pm0.03) \\ \\ =P(0.41\ \textless \ \hat{p}\ \textless \ 0.47)$

The standardization of the sample distribution of proportions is given by:

$P(a\ \textless \ \hat{p}\ \textless \ b)=P\left(\frac{a-p}{\sqrt{\frac{p(1-p)}{n}}}\ \textless \ z\ \textless \ \frac{b-p}{\sqrt{\frac{p(1-p)}{n}}}\right) \\ \\ =P\left(z\ \textless \ \frac{b-p}{\sqrt{\frac{p(1-p)}{n}}}\right)-P\left(\frac{a-p}{\sqrt{\frac{p(1-p)}{n}}}\right) \\ \\ \Rightarrow P(0.41\ \textless \ \hat{p}\ \textless \ 0.47)=P\left(\frac{0.47-0.44}{\sqrt{\frac{0.44(1-0.44)}{1400}}}\right)-P\left(\frac{0.41-0.44}{\sqrt{\frac{0.44(1-0.44)}{1400}}}\right)$
$\\ =P\left(\frac{0.03}{\sqrt{\frac{0.44(0.56)}{1400}}}\right)-P\left(\frac{-0.03}{\sqrt{\frac{0.44(0.56)}{1400}}}\right)=P\left(\frac{0.03}{\sqrt{\frac{0.2464}{1400}}}\right)-P\left(\frac{-0.03}{\sqrt{\frac{0.2464}{1400}}}\right) \\ \\ =P\left(\frac{0.03}{\sqrt{0.000176}}\right)-P\left(\frac{-0.03}{\sqrt{0.000176}}\right)=P\left(\frac{0.03}{0.013266}\right)-P\left(\frac{-0.03}{0.013266}\right) \\ \\ =P(2.26)-P(-2.26)=0.9881-0.0119=0.9762$