AP Statistics Flashcards: Confidence Interval For A Population Mean

Study Confidence Interval For A Population Mean in AP Statistics with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

AP Statistics

Confidence Interval For A Population Mean

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What happens to the interval if sample standard deviation decreases?

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ANSWER

The interval becomes narrower. Smaller ss reduces standard error and margin of error.

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Flashcard 1: What happens to the interval if sample standard deviation decreases?

Answer: The interval becomes narrower. Smaller ss reduces standard error and margin of error.

Flashcard 2: How does variability affect the confidence interval?

Answer: Greater variability widens the interval. Higher ss increases standard error and interval width.

Flashcard 3: Find tt^* for df=10df = 10 at 95% confidence level.

Answer: Approximately 2.228. From tt-table with df=10df = 10 and 95% confidence level.

Flashcard 4: Find the 95% confidence interval: xˉ=10\bar{x} = 10, t=2t^* = 2, s=5s = 5, n=25n = 25.

Answer: 10±210 \pm 2. Margin of error is 2×525=2×1=22 \times \frac{5}{\sqrt{25}} = 2 \times 1 = 2.

Flashcard 5: Find tt^* for df=10df = 10 at 95% confidence level.

Answer: Approximately 2.228. From tt-table with df=10df = 10 and 95% confidence level.

Flashcard 6: Identify the conditions for normality in constructing confidence intervals.

Answer: Sample size large or population distribution approximately normal. Ensures sampling distribution of xˉ\bar{x} is approximately normal.

Flashcard 7: Which option increases the precision of a confidence interval?

Answer: Increasing sample size. Larger sample reduces standard error and interval width.

Flashcard 8: Find the degrees of freedom for n=15n = 15.

Answer:

  1. Always n1n - 1 for single-sample tt-procedures.

Flashcard 9: What happens to a confidence interval if the confidence level decreases?

Answer: The interval becomes narrower. Lower confidence level uses smaller critical value.

Flashcard 10: What is the central limit theorem in context of confidence intervals?

Answer: The sampling distribution of the sample mean is approximately normal. Allows use of normal approximation for sample means.

Flashcard 11: Identify the degrees of freedom for a sample size of 25.

Answer:

  1. Degrees of freedom equals n1=251=24n - 1 = 25 - 1 = 24.

Flashcard 12: What does a 95% confidence level imply?

Answer: 95% of such intervals will capture the true population mean. Long-run frequency of intervals containing true parameter.

Flashcard 13: State the formula for a confidence interval for a mean.

Answer: xˉ±t(sn)\bar{x} \, \pm \, t^* \left(\frac{s}{\sqrt{n}}\right). Sample mean plus/minus margin of error using tt-distribution.

Flashcard 14: What is the relationship between confidence level and margin of error?

Answer: Higher confidence increases margin of error. Higher confidence requires larger critical value and margin.

Flashcard 15: State the formula for a confidence interval for a mean.

Answer: xˉ±t(sn)\bar{x} \, \pm \, t^{*} \left(\frac{s}{\sqrt{n}}\right). Sample mean plus/minus margin of error using tt-distribution.

Flashcard 16: Which table do you use to find tt^* values?

Answer: The tt-distribution table. Critical values depend on confidence level and degrees of freedom.

Flashcard 17: Find the 95% confidence interval: xˉ=10\bar{x} = 10, t=2t^* = 2, s=5s = 5, n=25n = 25.

Answer: 10±210 \pm 2. Margin of error is 2×525=2×1=22 \times \frac{5}{\sqrt{25}} = 2 \times 1 = 2.

Flashcard 18: How does variability affect the confidence interval?

Answer: Greater variability widens the interval. Higher ss increases standard error and interval width.

Flashcard 19: Identify the effect of decreasing variability on the confidence interval.

Answer: It narrows the interval. Less variability reduces standard error and margin of error.

Flashcard 20: Calculate the standard error: s=3s = 3, n=36n = 36.

Answer: 336=0.5\frac{3}{\sqrt{36}} = 0.5. Standard error formula: ss divided by square root of nn.

Flashcard 21: Find the margin of error given t=2t^* = 2, s=4s = 4, n=16n = 16.

Answer: 2×416=22 \times \frac{4}{\sqrt{16}} = 2. Margin of error equals t×t^* \times standard error.

Flashcard 22: What does tt^* represent in the confidence interval formula?

Answer: Critical value from the tt-distribution. Based on confidence level and degrees of freedom.

Flashcard 23: State the formula for standard error of the mean.

Answer: sn\frac{s}{\sqrt{n}}. Standard deviation of the sampling distribution of xˉ\bar{x}.

Flashcard 24: What is the relationship between confidence level and margin of error?

Answer: Higher confidence increases margin of error. Higher confidence requires larger critical value and margin.

Flashcard 25: What does xˉ\bar{x} represent in the confidence interval formula?

Answer: Sample mean. The average of all observations in the sample.

Flashcard 26: State the formula for standard error of the mean.

Answer: sn\frac{s}{\sqrt{n}}. Standard deviation of the sampling distribution of xˉ\bar{x}.

Flashcard 27: What is the effect of confidence level on interval width?

Answer: Higher confidence level widens the interval. More confidence requires wider interval to capture parameter.

Flashcard 28: Calculate the standard error: s=3s = 3, n=36n = 36.

Answer: 336=0.5\frac{3}{\sqrt{36}} = 0.5. Standard error formula: ss divided by square root of nn.

Flashcard 29: Calculate confidence interval: xˉ=20\bar{x} = 20, t=1.96t^* = 1.96, s=4s = 4, n=100n = 100.

Answer: 20±0.78420 \pm 0.784. Margin of error: 1.96×4100=1.96×0.4=0.7841.96 \times \frac{4}{\sqrt{100}} = 1.96 \times 0.4 = 0.784.

Flashcard 30: What is the effect of confidence level on interval width?

Answer: Higher confidence level widens the interval. More confidence requires wider interval to capture parameter.

Flashcard 31: What does xˉ\bar{x} represent in the confidence interval formula?

Answer: Sample mean. The average of all observations in the sample.

Flashcard 32: How do you interpret a confidence interval?

Answer: A range where the population mean is likely to be found. Plausible values for the unknown population mean.

Flashcard 33: Identify the condition for using a tt-distribution in confidence intervals.

Answer: Population standard deviation is unknown and sample is small. Use when σ\sigma is unknown, especially with small samples.

Flashcard 34: Which distribution is used when population standard deviation is known?

Answer: Normal distribution (Z-distribution). Known σ\sigma allows use of standard normal distribution.

Flashcard 35: Identify the sample mean in the data set: 5, 6, 7, 8, 9.

Answer: xˉ=7\bar{x} = 7. Sum of values divided by count: (5+6+7+8+9)/5=7(5+6+7+8+9)/5 = 7.

Flashcard 36: How does sample size relate to the standard error?

Answer: Larger sample size decreases standard error. Standard error decreases as nn increases: sn\frac{s}{\sqrt{n}}.

Flashcard 37: Calculate the interval: xˉ=50\bar{x} = 50, t=2.5t^* = 2.5, s=10s = 10, n=16n = 16.

Answer: 50±6.2550 \pm 6.25. Margin of error: 2.5×1016=2.5×2.5=6.252.5 \times \frac{10}{\sqrt{16}} = 2.5 \times 2.5 = 6.25.

Flashcard 38: Which term describes the range of values in a confidence interval?

Answer: Margin of error. Half-width of confidence interval around point estimate.

Flashcard 39: What does ss represent in the confidence interval formula?

Answer: Sample standard deviation. Measures spread of sample observations around sample mean.

Flashcard 40: Which distribution is used when population standard deviation is known?

Answer: Normal distribution (Z-distribution). Known σ\sigma allows use of standard normal distribution.

Flashcard 41: Which factor has no effect on interval width if other factors are fixed?

Answer: Sample mean. Sample mean is center point; doesn't affect interval width.

Flashcard 42: Find the margin of error given t=2t^* = 2, s=4s = 4, n=16n = 16.

Answer: 2×416=22 \times \frac{4}{\sqrt{16}} = 2. Margin of error equals t×t^* \times standard error.

Flashcard 43: What does nn represent in the confidence interval formula?

Answer: Sample size. Total number of observations in the sample.

Flashcard 44: What does a 95% confidence level imply?

Answer: 95% of such intervals will capture the true population mean. Long-run frequency of intervals containing true parameter.

Flashcard 45: Identify the conditions for normality in constructing confidence intervals.

Answer: Sample size large or population distribution approximately normal. Ensures sampling distribution of xˉ\bar{x} is approximately normal.

Flashcard 46: Identify the effect of increasing sample size on the confidence interval.

Answer: It narrows the confidence interval. Larger nn reduces standard error and margin of error.

Flashcard 47: What is the central limit theorem in context of confidence intervals?

Answer: The sampling distribution of the sample mean is approximately normal. Allows use of normal approximation for sample means.

Flashcard 48: Identify the effect of decreasing variability on the confidence interval.

Answer: It narrows the interval. Less variability reduces standard error and margin of error.

Flashcard 49: Which factor has no effect on interval width if other factors are fixed?

Answer: Sample mean. Sample mean is center point; doesn't affect interval width.

Flashcard 50: Identify the sample mean in the data set: 5, 6, 7, 8, 9.

Answer: xˉ=7\bar{x} = 7. Sum of values divided by count: (5+6+7+8+9)/5=7(5+6+7+8+9)/5 = 7.

Flashcard 51: Find the degrees of freedom for n=15n = 15.

Answer:

  1. Always n1n - 1 for single-sample tt-procedures.

Flashcard 52: What does a wider confidence interval suggest about precision?

Answer: Less precision. Wider intervals indicate greater uncertainty about parameter.

Flashcard 53: What does ss represent in the confidence interval formula?

Answer: Sample standard deviation. Measures spread of sample observations around sample mean.

Flashcard 54: Identify the degrees of freedom for a sample size of 25.

Answer:

  1. Degrees of freedom equals n1=251=24n - 1 = 25 - 1 = 24.

Flashcard 55: What is the primary purpose of a confidence interval?

Answer: To estimate a population parameter. Uses sample data to infer about population parameters.