Q: Given the data set 8, 10 , 7 , 2 , 6 , 9 , 3 , 8 , 9. What is the median? 7 10 9 8

The correct answer is: $8$ Explanation We have 9 data points. First sort them in ascending order: $2,3,6,7,8,8,9,9,10$. Steps: Count the number of observations: $n=9$. Find the middle position using $$\frac{n+1}{2}$$ which gives $$\frac{9+1}{2}=5.$$ The median is the 5th value in the sorted list, which is $8$. Therefore, the median is $8$.

Q: 5'4 in Meters

The correct answer is: $1.6256\text{ m}$ (about $1.63\text{ m}$) Explanation 5'4" means 5 feet and 4 inches. Use the exact metric conversions $1\text{ ft}=0.3048\text{ m}$ and $1\text{ in}=0.0254\text{ m}$. Steps: Convert feet: $$5\times0.3048=1.524\text{ m}$$ Convert inches: $$4\times0.0254=0.1016\text{ m}$$ Add: $$1.524+0.1016=1.6256\text{ m}$$ Rounded to two decimal places: $1.63\text{ m}$.

Q: This table shows how many male and female students attended two different movies. What is the probability that a randomly chosen person from this group is male and attended an action movie? Round your answer to two decimal places. Action Drama Total Male 105 124 229 Female 99 151 250 Total 204 275 479 A. 0.22 B. 0.52 C. 0.43 D. 0.11

The correct answer is A. Explanation We want the probability a randomly chosen person is male and attended an action movie. That count is 105 out of the total 479 people, so Compute the probability: $P=\dfrac{105}{479}$ Convert to decimal: $P\approx 0.21920668$ Round to two decimal places: $0.22$ Therefore the correct choice is A (0.22).

Question

Calculate and interpret a 95% confidence interval for the proportion of all Americans who would report they "often or always" got enough sleep during the last 7 nights.

Accepted Answer

I can’t compute the 95% confidence interval without the sample size and the number (or sample proportion) who said “often or always.” Please provide those. Meanwhile, here’s the formula, steps, a worked example, and how to interpret the result.

Explanation

Formula for a 95% CI for a population proportion:$$\hat{p}\pm z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$where $z^*=1.96$ for a 95% CI, $\hat{p}$ is the sample proportion, and $n$ is the sample size.

Conditions: the sample should be random (or representative), and the normal approximation is OK if $n\hat{p}\ge10$ and $n(1-\hat{p})\ge10$.

Steps:

Compute the sample proportion: $\hat{p}=\dfrac{	ext{number who said "often or always"}}{n}$.
Compute the standard error: $\mathrm{SE}=\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$.
Compute margin of error: $ME=1.96	imes\mathrm{SE}$.
CI is $\hat{p}\pm ME$ (report as a percentage if you prefer).

Worked example (hypothetical): suppose $n=1000$ and 360 respondents said “often or always,” so $\hat{p}=360/1000=0.36$.

$\mathrm{SE}=\sqrt{\dfrac{0.36(1-0.36)}{1000}}=\sqrt{\dfrac{0.2304}{1000}}=\sqrt{0.0002304}=0.01518$.
$ME=1.96	imes0.01518=0.0298$.
95% CI $=0.36\pm0.0298=(0.3302,\;0.3898)$, i.e. about (33.0%, 39.0%).

Interpretation: We are 95% confident that the true proportion of all Americans who would report they “often or always” got enough sleep in the past 7 nights lies between about 33.0% and 39.0%. (This means the method used to produce the interval will capture the true proportion in about 95% of repeated random samples of the same size; it is not a 95% probability for the specific interval.)

If you give me the sample size and the count (or sample proportion), I’ll compute the exact 95% CI for your data.

Calculate and interpret a 95% confidence interval for the proportion of all Americans who would report they “often or always” got enough sleep during the last 7 nights.

Explanation