Estimating Population Parameters with Error Margin - Statistics
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What is the statistic used to estimate $\mu$ in a sample survey?
What is the statistic used to estimate $\mu$ in a sample survey?
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Sample mean $\bar{x}$. Sample mean $\bar{x}$ estimates population mean $\mu$.
Sample mean $\bar{x}$. Sample mean $\bar{x}$ estimates population mean $\mu$.
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What is the point estimate for a population proportion $p$ from a sample of size $n$ with $x$ successes?
What is the point estimate for a population proportion $p$ from a sample of size $n$ with $x$ successes?
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$\hat{p}=\frac{x}{n}$. Sample proportion equals successes divided by sample size.
$\hat{p}=\frac{x}{n}$. Sample proportion equals successes divided by sample size.
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What is the margin of error if a simulation gives a central $95%$ interval $[L,U]$ for a parameter?
What is the margin of error if a simulation gives a central $95%$ interval $[L,U]$ for a parameter?
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$\text{ME}=\frac{U-L}{2}$. ME is half the width of the confidence interval.
$\text{ME}=\frac{U-L}{2}$. ME is half the width of the confidence interval.
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What is the point estimate for a population mean $\mu$ based on a sample mean $\bar{x}$?
What is the point estimate for a population mean $\mu$ based on a sample mean $\bar{x}$?
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$\bar{x}$. Sample mean directly estimates population mean.
$\bar{x}$. Sample mean directly estimates population mean.
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Identify the parameter: A poll estimates the fraction of all voters who support a candidate.
Identify the parameter: A poll estimates the fraction of all voters who support a candidate.
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Population proportion $p$. Fraction of all voters is a population proportion.
Population proportion $p$. Fraction of all voters is a population proportion.
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Identify the parameter: A survey estimates the average number of hours students sleep per night.
Identify the parameter: A survey estimates the average number of hours students sleep per night.
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Population mean $\mu$. Average for all students is a population mean.
Population mean $\mu$. Average for all students is a population mean.
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Which sampling method best justifies inference to a population: convenience sample or random sample?
Which sampling method best justifies inference to a population: convenience sample or random sample?
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Random sample. Random sampling ensures representativeness for inference.
Random sample. Random sampling ensures representativeness for inference.
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Which condition is needed so simulation-based inference targets the population: random sampling or voluntary response?
Which condition is needed so simulation-based inference targets the population: random sampling or voluntary response?
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Random sampling. Random sampling makes sample representative of population.
Random sampling. Random sampling makes sample representative of population.
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Find $\hat{p}$: In a random sample of $n=200$, $x=56$ people answer "yes".
Find $\hat{p}$: In a random sample of $n=200$, $x=56$ people answer "yes".
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$\hat{p}=\frac{56}{200}=0.28$. Divide successes by sample size: $\frac{56}{200}=0.28$.
$\hat{p}=\frac{56}{200}=0.28$. Divide successes by sample size: $\frac{56}{200}=0.28$.
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Find the estimate with ME: If $\hat{p}=0.28$ and $\text{ME}=0.04$, what is the interval estimate?
Find the estimate with ME: If $\hat{p}=0.28$ and $\text{ME}=0.04$, what is the interval estimate?
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$0.28\pm^0.04=[0.24,0.32]$. Point estimate $\pm$ ME gives confidence interval.
$0.28\pm^0.04=[0.24,0.32]$. Point estimate $\pm$ ME gives confidence interval.
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Find the estimate with ME: If $\bar{x}=72$ and $\text{ME}=3$, what is the interval estimate for $\mu$?
Find the estimate with ME: If $\bar{x}=72$ and $\text{ME}=3$, what is the interval estimate for $\mu$?
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$72\pm^3=[69,75]$. Point estimate $\pm$ ME gives confidence interval.
$72\pm^3=[69,75]$. Point estimate $\pm$ ME gives confidence interval.
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Find the margin of error: A central $95%$ simulation interval for $p$ is $[0.41,0.49]$.
Find the margin of error: A central $95%$ simulation interval for $p$ is $[0.41,0.49]$.
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$\text{ME}=\frac{0.49-0.41}{2}=0.04$. ME equals half the interval width: $\frac{0.08}{2}=0.04$.
$\text{ME}=\frac{0.49-0.41}{2}=0.04$. ME equals half the interval width: $\frac{0.08}{2}=0.04$.
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Find the point estimate: A central $95%$ simulation interval for $\mu$ is $[18.2,19.0]$.
Find the point estimate: A central $95%$ simulation interval for $\mu$ is $[18.2,19.0]$.
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$\text{point estimate}=\frac{18.2+19.0}{2}=18.6$. Point estimate is interval midpoint: $\frac{37.2}{2}=18.6$.
$\text{point estimate}=\frac{18.2+19.0}{2}=18.6$. Point estimate is interval midpoint: $\frac{37.2}{2}=18.6$.
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What does a $95%$ margin of error claim describe: variability from random sampling or individual accuracy?
What does a $95%$ margin of error claim describe: variability from random sampling or individual accuracy?
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Variability from random sampling. ME quantifies sampling variability, not individual errors.
Variability from random sampling. ME quantifies sampling variability, not individual errors.
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In simulation for a sample proportion, what quantity is repeatedly recomputed each trial to form a sampling distribution?
In simulation for a sample proportion, what quantity is repeatedly recomputed each trial to form a sampling distribution?
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$\hat{p}$ from each simulated sample. Each trial computes $\hat{p}$ to build sampling distribution.
$\hat{p}$ from each simulated sample. Each trial computes $\hat{p}$ to build sampling distribution.
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In simulation for a sample mean, what quantity is repeatedly recomputed each trial to form a sampling distribution?
In simulation for a sample mean, what quantity is repeatedly recomputed each trial to form a sampling distribution?
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$\bar{x}$ from each simulated sample. Each trial computes $\bar{x}$ to build sampling distribution.
$\bar{x}$ from each simulated sample. Each trial computes $\bar{x}$ to build sampling distribution.
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Which change usually decreases simulation-based margin of error: increasing $n$ or decreasing $n$?
Which change usually decreases simulation-based margin of error: increasing $n$ or decreasing $n$?
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Increasing $n$. Larger samples reduce sampling variability and ME.
Increasing $n$. Larger samples reduce sampling variability and ME.
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Choose the correct interpretation: "$\hat{p}=0.52\pm^0.03$" refers to $p$ or to the sample proportion?
Choose the correct interpretation: "$\hat{p}=0.52\pm^0.03$" refers to $p$ or to the sample proportion?
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It estimates the population proportion $p$. Interval estimates target the population parameter $p$.
It estimates the population proportion $p$. Interval estimates target the population parameter $p$.
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What is the statistic used to estimate $p$ in a sample survey?
What is the statistic used to estimate $p$ in a sample survey?
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Sample proportion $\hat{p}$. Sample proportion $\hat{p}$ estimates population proportion $p$.
Sample proportion $\hat{p}$. Sample proportion $\hat{p}$ estimates population proportion $p$.
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What is the estimate written using point estimate and margin of error?
What is the estimate written using point estimate and margin of error?
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$\text{estimate}=\text{point estimate}\pm\text{ME}$. Interval estimate combines point estimate with margin of error.
$\text{estimate}=\text{point estimate}\pm\text{ME}$. Interval estimate combines point estimate with margin of error.
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Find $\hat{p}$ when a survey has $x=42$ successes out of $n=120$ respondents.
Find $\hat{p}$ when a survey has $x=42$ successes out of $n=120$ respondents.
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$\hat{p}=\frac{42}{120}=0.35$. Applies formula: $\frac{42}{120}=0.35$.
$\hat{p}=\frac{42}{120}=0.35$. Applies formula: $\frac{42}{120}=0.35$.
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What is the point estimate for a population mean $\mu$ from a sample with values $x_1,\dots,x_n$?
What is the point estimate for a population mean $\mu$ from a sample with values $x_1,\dots,x_n$?
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$\bar{x}=\frac{\sum_{i=1}^{n} x_i}{n}$. Sum all values and divide by sample size.
$\bar{x}=\frac{\sum_{i=1}^{n} x_i}{n}$. Sum all values and divide by sample size.
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Compute the $95%$ rule-of-thumb MOE for a mean when $s=12$ and $n=36$.
Compute the $95%$ rule-of-thumb MOE for a mean when $s=12$ and $n=36$.
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$MOE\approx 2\frac{12}{\sqrt{36}}=4$. Applies rule: $2\times\frac{12}{6}=4$.
$MOE\approx 2\frac{12}{\sqrt{36}}=4$. Applies rule: $2\times\frac{12}{6}=4$.
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A simulation gives the middle $95%$ of $\hat{p}$ as $[0.41,0.49]$. What is the simulation-based $MOE$?
A simulation gives the middle $95%$ of $\hat{p}$ as $[0.41,0.49]$. What is the simulation-based $MOE$?
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$MOE\approx\frac{0.49-0.41}{2}=0.04$. Half the interval width: $\frac{0.08}{2}=0.04$.
$MOE\approx\frac{0.49-0.41}{2}=0.04$. Half the interval width: $\frac{0.08}{2}=0.04$.
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Identify the parameter estimated by $\hat{p}$ in a sample survey.
Identify the parameter estimated by $\hat{p}$ in a sample survey.
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The population proportion $p$. Sample proportion estimates true population proportion.
The population proportion $p$. Sample proportion estimates true population proportion.
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Identify the parameter estimated by $\bar{x}$ in a sample survey.
Identify the parameter estimated by $\bar{x}$ in a sample survey.
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The population mean $\mu$. Sample mean estimates true population mean.
The population mean $\mu$. Sample mean estimates true population mean.
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What is the key idea of using simulation to get a margin of error for random sampling?
What is the key idea of using simulation to get a margin of error for random sampling?
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Use the variability of simulated sample statistics. Simulated spread shows sampling uncertainty.
Use the variability of simulated sample statistics. Simulated spread shows sampling uncertainty.
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In a simulation-based MOE, which quantity is typically used as MOE from the simulated sampling distribution?
In a simulation-based MOE, which quantity is typically used as MOE from the simulated sampling distribution?
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About the middle $95%$ half-width (e.g., $97.5%-2.5%$)/$2$. Half-width of middle 95% gives MOE.
About the middle $95%$ half-width (e.g., $97.5%-2.5%$)/$2$. Half-width of middle 95% gives MOE.
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What is a margin of error (MOE) in an estimate written as estimate $\pm$ MOE?
What is a margin of error (MOE) in an estimate written as estimate $\pm$ MOE?
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Half the width of the interval around the estimate. MOE is the plus/minus value from the center estimate.
Half the width of the interval around the estimate. MOE is the plus/minus value from the center estimate.
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Which sampling method best justifies inference to a population: random sample or voluntary response?
Which sampling method best justifies inference to a population: random sample or voluntary response?
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Random sample. Random sampling ensures unbiased representation.
Random sample. Random sampling ensures unbiased representation.
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