Statistics Flashcards: Estimating Population Parameters With Error Margin

Study Estimating Population Parameters With Error Margin in Statistics with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

Statistics

Estimating Population Parameters With Error Margin

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Choose the correct interpretation: "p^=0.52±0.03\hat{p}=0.52\pm^0.03" refers to pp or to the sample proportion?

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ANSWER

It estimates the population proportion pp. Interval estimates target the population parameter pp.

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Flashcard 1: Choose the correct interpretation: "p^=0.52±0.03\hat{p}=0.52\pm^0.03" refers to pp or to the sample proportion?

Answer: It estimates the population proportion pp. Interval estimates target the population parameter pp.

Flashcard 2: What is the margin of error if a simulation gives a central 95%95\% interval [L,U][L,U] for a parameter?

Answer: ME=UL2\text{ME}=\frac{U-L}{2}. ME is half the width of the confidence interval.

Flashcard 3: Which sampling method best justifies inference to a population: convenience sample or random sample?

Answer: Random sample. Random sampling ensures representativeness for inference.

Flashcard 4: What is the key idea of using simulation to get a margin of error for random sampling?

Answer: Use the variability of simulated sample statistics. Simulated spread shows sampling uncertainty.

Flashcard 5: What is the approximate 95%95\% margin of error for a mean using the rule of thumb?

Answer: MOE2snMOE\approx 2\frac{s}{\sqrt{n}}. Uses z2z\approx 2 for 95% confidence with mean SE.

Flashcard 6: Identify the parameter: A survey estimates the average number of hours students sleep per night.

Answer: Population mean μ\mu. Average for all students is a population mean.

Flashcard 7: What is the point estimate for a population mean μ\mu from a sample with values x1,,xnx_1,\dots,x_n?

Answer: xˉ=i=1nxin\bar{x}=\frac{\sum_{i=1}^{n} x_i}{n}. Sum all values and divide by sample size.

Flashcard 8: What is the statistic used to estimate pp in a sample survey?

Answer: Sample proportion p^\hat{p}. Sample proportion p^\hat{p} estimates population proportion pp.

Flashcard 9: What is the point estimate for a population mean μ\mu based on a sample mean xˉ\bar{x}?

Answer: xˉ\bar{x}. Sample mean directly estimates population mean.

Flashcard 10: Find the point estimate: A central 95%95\% simulation interval for μ\mu is [18.2,19.0][18.2,19.0].

Answer: point estimate=18.2+19.02=18.6\text{point estimate}=\frac{18.2+19.0}{2}=18.6. Point estimate is interval midpoint: 37.22=18.6\frac{37.2}{2}=18.6.

Flashcard 11: In a simulation-based MOE, which quantity is typically used as MOE from the simulated sampling distribution?

Answer: About the middle 95%95\% half-width (e.g., 97.5%2.5%97.5\%-2.5\%)/22. Half-width of middle 95% gives MOE.

Flashcard 12: Which sampling method best justifies inference to a population: random sample or voluntary response?

Answer: Random sample. Random sampling ensures unbiased representation.

Flashcard 13: Find p^\hat{p}: In a random sample of n=200n=200, x=56x=56 people answer "yes".

Answer: p^=56200=0.28\hat{p}=\frac{56}{200}=0.28. Divide successes by sample size: 56200=0.28\frac{56}{200}=0.28.

Flashcard 14: Find the estimate with ME: If xˉ=72\bar{x}=72 and ME=3\text{ME}=3, what is the interval estimate for μ\mu?

Answer: 72±3=[69,75]72\pm^3=[69,75]. Point estimate ±\pm ME gives confidence interval.

Flashcard 15: What is the standard error for a sample proportion in terms of p^\hat{p} and nn?

Answer: SEp^=p^(1p^)nSE_{\hat{p}}=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}. Uses proportion and its complement divided by nn.

Flashcard 16: Identify the parameter estimated by p^\hat{p} in a sample survey.

Answer: The population proportion pp. Sample proportion estimates true population proportion.

Flashcard 17: What is the point estimate for a population proportion pp from a sample of size nn with xx successes?

Answer: p^=xn\hat{p}=\frac{x}{n}. Sample proportion equals successes divided by sample size.

Flashcard 18: Identify the parameter: A poll estimates the fraction of all voters who support a candidate.

Answer: Population proportion pp. Fraction of all voters is a population proportion.

Flashcard 19: Find the estimate with ME: If p^=0.28\hat{p}=0.28 and ME=0.04\text{ME}=0.04, what is the interval estimate?

Answer: 0.28±0.04=[0.24,0.32]0.28\pm^0.04=[0.24,0.32]. Point estimate ±\pm ME gives confidence interval.

Flashcard 20: Find p^\hat{p} when a survey has x=42x=42 successes out of n=120n=120 respondents.

Answer: p^=42120=0.35\hat{p}=\frac{42}{120}=0.35. Applies formula: 42120=0.35\frac{42}{120}=0.35.

Flashcard 21: What is the general form of a confidence interval for a parameter using a critical value?

Answer: Estimate ±\pm (critical value)×\times(standard error). Multiplies critical value by SE for interval width.

Flashcard 22: What is a margin of error (MOE) in an estimate written as estimate ±\pm MOE?

Answer: Half the width of the interval around the estimate. MOE is the plus/minus value from the center estimate.

Flashcard 23: What is the standard error for a sample mean in terms of sample standard deviation ss and nn?

Answer: SExˉ=snSE_{\bar{x}}=\frac{s}{\sqrt{n}}. Divides sample SD by square root of sample size.

Flashcard 24: What is the statistic used to estimate μ\mu in a sample survey?

Answer: Sample mean xˉ\bar{x}. Sample mean xˉ\bar{x} estimates population mean μ\mu.

Flashcard 25: What is the general form of an estimate with margin of error for a parameter?

Answer: Estimate ±\pm margin of error. Standard format shows uncertainty range.

Flashcard 26: What is the estimate written using point estimate and margin of error?

Answer: estimate=point estimate±ME\text{estimate}=\text{point estimate}\pm\text{ME}. Interval estimate combines point estimate with margin of error.

Flashcard 27: Identify the parameter estimated by xˉ\bar{x} in a sample survey.

Answer: The population mean μ\mu. Sample mean estimates true population mean.

Flashcard 28: A simulation gives the middle 95%95\% of p^\hat{p} as [0.41,0.49][0.41,0.49]. What is the simulation-based MOEMOE?

Answer: MOE0.490.412=0.04MOE\approx\frac{0.49-0.41}{2}=0.04. Half the interval width: 0.082=0.04\frac{0.08}{2}=0.04.

Flashcard 29: What does a 95%95\% margin of error claim describe: variability from random sampling or individual accuracy?

Answer: Variability from random sampling. ME quantifies sampling variability, not individual errors.

Flashcard 30: In simulation for a sample proportion, what quantity is repeatedly recomputed each trial to form a sampling distribution?

Answer: p^\hat{p} from each simulated sample. Each trial computes p^\hat{p} to build sampling distribution.

Flashcard 31: What is the approximate 95%95\% margin of error for a proportion using the rule of thumb?

Answer: MOE2p^(1p^)nMOE\approx 2\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}. Uses z2z\approx 2 for 95% confidence with proportion SE.

Flashcard 32: Which condition is needed so simulation-based inference targets the population: random sampling or voluntary response?

Answer: Random sampling. Random sampling makes sample representative of population.

Flashcard 33: Which change usually decreases simulation-based margin of error: increasing nn or decreasing nn?

Answer: Increasing nn. Larger samples reduce sampling variability and ME.

Flashcard 34: Find the margin of error: A central 95%95\% simulation interval for pp is [0.41,0.49][0.41,0.49].

Answer: ME=0.490.412=0.04\text{ME}=\frac{0.49-0.41}{2}=0.04. ME equals half the interval width: 0.082=0.04\frac{0.08}{2}=0.04.

Flashcard 35: In simulation for a sample mean, what quantity is repeatedly recomputed each trial to form a sampling distribution?

Answer: xˉ\bar{x} from each simulated sample. Each trial computes xˉ\bar{x} to build sampling distribution.

Flashcard 36: Compute the 95%95\% rule-of-thumb MOE for a mean when s=12s=12 and n=36n=36.

Answer: MOE21236=4MOE\approx 2\frac{12}{\sqrt{36}}=4. Applies rule: 2×126=42\times\frac{12}{6}=4.