Question

1. - / 4 points a sample of size 110 is selected from a population with p = 0.50. (a) what is the expected value of \\(\hat{p}\\)? (b) what is the standard error of \\(\hat{p}\\)? (round your answer to four decimal places.) (c) show the sampling distribution of \\(\hat{p}\\). (d) what does the sampling distribution of \\(\hat{p}\\) show? \\(\circ\\) the probability distribution of \\(\hat{p}\\) is approximately normal with a mean of 110. \\(\circ\\) the probability distribution of \\(\hat{p}\\) is approximately normal with a mean of 0.0477. \\(\circ\\) the probability distribution of \\(\hat{p}\\) is approximately normal with a mean of 0.50. \\(\circ\\) the probability distribution of \\(\hat{p}\\) is approximately normal with a mean of 0. 2. - / 5 points

Explanation:

Part (a)

Step1: Recall formula for expected value of $\hat{p}$

The expected value of the sample proportion $\hat{p}$ (also called the mean of the sampling distribution of $\hat{p}$) is equal to the population proportion $p$. So the formula is $E(\hat{p}) = p$.

Step2: Substitute the given value of $p$

We are given that $p = 0.50$. So substituting into the formula, we get $E(\hat{p}) = 0.50$.

Step1: Recall formula for standard error of $\hat{p}$

The formula for the standard error (SE) of the sample proportion $\hat{p}$ is $SE = \sqrt{\frac{p(1 - p)}{n}}$, where $p$ is the population proportion and $n$ is the sample size.

Step2: Substitute the given values

We know that $p = 0.50$, $1 - p = 1 - 0.50 = 0.50$, and $n = 110$. Substituting these values into the formula, we get $SE = \sqrt{\frac{0.50\times0.50}{110}}$.

Step3: Calculate the value

First, calculate the numerator: $0.50\times0.50 = 0.25$. Then divide by $n$: $\frac{0.25}{110}\approx0.00227273$. Then take the square root: $\sqrt{0.00227273}\approx0.04767$. Rounding to four decimal places, we get $0.0477$.

We know from the properties of the sampling distribution of the sample proportion $\hat{p}$ that if $np\geq5$ and $n(1 - p)\geq5$ (here $np = 110\times0.50 = 55\geq5$ and $n(1 - p)=110\times0.50 = 55\geq5$), the sampling distribution of $\hat{p}$ is approximately normal with mean equal to the population proportion $p$. From part (a), the mean of $\hat{p}$ is $0.50$.

• The first option is wrong because the mean of $\hat{p}$ is not the sample size $n = 110$.
• The second option is wrong because the mean is $0.50$, not $0.0477$ (which is the standard error).
• The third option is correct as the mean of the sampling distribution of $\hat{p}$ is $p = 0.50$ and the distribution is approximately normal.
• The fourth option is wrong because the mean is $0.50$, not $0$.

Answer:

$0.50$

Part (b)