Question

The United States government wanted to determine what proportion of Americans approve of the current President so they conducted a survey of 2,000 randomly chosen Americans. Assume that the true proportion of Americans who approve of the president is 0.46. Determine what the mean and the standard deviation of the sampling distribution of the sample proportion would be. Round to 2 decimal places when necessary.

Answer 1

Answer:

$E(\hat p) = \mu_{\hat p} =p =0.46$

$Sd(\hat p) = \sigma_{\hat p}= \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.46*(1-0.46)}{2000}}= 0.011 \approx 0.01$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

For this case we can assume  that the distribution for the sample proportion is given by:

$\hat p \sim N (p,\sqrt{\frac{p(1-p)}{n}})$

Where the mean is given by:

$E(\hat p) = \mu_{\hat p} =p =0.46$

And the standard deviation is given by:

$Sd(\hat p) = \sigma_{\hat p}= \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.46*(1-0.46)}{2000}}= 0.011 \approx 0.01$