Question

Suppose 60% of American adults believe Martha Stewart is guilty of obstruction of justice and fraud related to insider trading. We will take a random sample of 100 American adults and ask them the question. Then the sampling distribution of the sample proportion of people who answer yes to the question is:a. neither Normal, not Binomial.b. approximately Normal, with unknown mean and standard deviation.c. approximately Normal, with mean 0.6 and standard error 0.04899.d. Binomial, with n=100 and p=0.60.

Answer 1

Answer:

Since the sample size is large enough (n>30) and the probability of success is near to 0.5, and we have that $n\hat p = 60>10$ and $n(1-\hat p) = 40>10$ we can assume that the distribution for $\hat p$ is normal

The population proportion have the following distribution  

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

The mean is given by:

$\mu_p = 0.6$

$\sigma_p = \sqrt{\frac{0.6*(1-0.6)}{100}}=0.04899$

So then we can conclude that the best answer would be:

c. approximately Normal, with mean 0.6 and standard error 0.04899

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".  

$p$ represent the real population proportion of interest

$\hat p$ represent the estimated proportion for the sample

n is the sample size required (variable of interest)

Solution to the problem

Since the sample size is large enough (n>30) and the probability of success is near to 0.5, and we have that $n\hat p = 60>10$ and $n(1-\hat p) = 40>10$ we can assume that the distribution for $\hat p$ is normal

The population proportion have the following distribution  

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

The mean is given by:

$\mu_p = 0.6$

$\sigma_p = \sqrt{\frac{0.6*(1-0.6)}{100}}=0.04899$

So then we can conclude that the best answer would be:

c. approximately Normal, with mean 0.6 and standard error 0.04899

Answer 2

The best answer would be c. approximately Normal, with a mean of 0.6 and standard error of 0.04899.

How to find the sample standard deviation for distribution of a sample proportion?

Suppose the sample proportion is denoted by $\hat{p}$ , then, its distribution is normally distributed with the  mean $\mu = p$ and the standard deviation of the distribution of $\hat{p}$ is given by

$\sigma = \sqrt{\dfrac{p(1-p)}{n}}$

where n is the sample size. It is true until

$np \geq 10\\and \\ n(1-p) \geq 10$

The population proportion has the following distribution  

$\sigma = \sqrt{\dfrac{p(1-p)}{n}}\\\\\sigma = \sqrt{\dfrac{0.6(1-0/.6)}{100}}$

 

= 0.04899

So then we can conclude that the best answer would be: c. approximately Normal, with a mean of 0.6 and standard error of 0.04899.

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