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AP Statistics : Confidence Intervals and Regression

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Example Questions

Example Question #1 : How To Find Confidence Intervals For The Slope Of A Least Squares Regression Line

You estimate a regression model with $\beta = 3.4$ $\displaystyle \beta = 3.4$ and $\displaystyle SE = .024$ , where $\beta$ $\displaystyle \beta$ is the beta coefficient and $\displaystyle SE$ is the standard error. Construct 95% confidence intervals for $\beta$ $\displaystyle \beta$ .

Possible Answers:

$\left [ 3.35,3.45 \right ]$ $\displaystyle \left [ 3.35,3.45 \right ]$

$\left [ 3.1,3.45 \right ]$ $\displaystyle \left [ 3.1,3.45 \right ]$

$\left [ 3.4,3.45 \right ]$ $\displaystyle \left [ 3.4,3.45 \right ]$

Correct answer:

$\left [ 3.35,3.45 \right ]$ $\displaystyle \left [ 3.35,3.45 \right ]$

Explanation:

To construct 95% confidence intervals for $\beta$ $\displaystyle \beta$ , we simply take the coefficient and add/subtract $1.96\times the\ standard\ error\ of\ \beta$ $\displaystyle 1.96\times the\ standard\ error\ of\ \beta$ . This is because $\beta$ $\displaystyle \beta$ is assumed to follow a symmetrical distribution (the normal), and 95% of the values in the sampling distribution are contained within 1.96 standard errors of $\beta$ $\displaystyle \beta$ .

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Example Question #1 : How To Find Confidence Intervals

Which of the following statements are correct about confidence intervals?

Possible Answers:

The width of a confidence interval increases as the sample size increases and increases as the confidence level decreases.

The width of a confidence interval does not change as the sample size increases and increases as the confidence level increases.

The width of a confidence interval decreases as the sample size increases and increases as the confidence level increases.

The width of a confidence interval increases as the sample size increases and increases as the confidence level increases.

The width of a confidence interval decreases as the sample size increases and increases as the confidence level decreases.

Correct answer:

The width of a confidence interval decreases as the sample size increases and increases as the confidence level increases.

Explanation:

Larger samples give narrower intervals. We are able to estimate a population proportion more precisely with a larger sample size.

As the confidence level increases the width of the confidence interval also increases. A larger confidence level increases the chance that the correct value will be found in the confidence interval. This means that the interval is larger.

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Example Question #12 : Confidence Intervals

You are asked to create a $95\%$ $\displaystyle 95\%$ confidence interval with a margin of error no larger than $\displaystyle 3$ while sampling from a normally distributed population with a standard deviation of $\displaystyle 5$ . What is the minimum required sample size?

Possible Answers:

$\displaystyle 11$

$\displaystyle 14$

$\displaystyle 13$

$\displaystyle 10$

$\displaystyle 12$

Correct answer:

$\displaystyle 11$

Explanation:

Keep in mind that the margin of error for a confidence interval based on a normal population is equal to $z^{*}\left(\frac{\sigma}{\sqrt{n}}\right)$ $\displaystyle z^{*}\left(\frac{\sigma}{\sqrt{n}}\right)$ , where $z^{*}$ $\displaystyle z^{*}$ is the $\displaystyle z$ -score corresponding to the desired confidence level.

From the problem, we can tell that $z^{*} = 1.96$ $\displaystyle z^{*} = 1.96$ and $\sigma = 5$ $\displaystyle \sigma = 5$ . We can then solve for $\displaystyle n$ algebraically:

$3 = 1.96\left(\frac{5}{\sqrt{n}}\right)$ $\displaystyle 3 = 1.96\left(\frac{5}{\sqrt{n}}\right)$

$\frac{3}{1.96} = \frac{5}{\sqrt{n}}$ $\displaystyle \frac{3}{1.96} = \frac{5}{\sqrt{n}}$

$\frac{3}{1.96}\cdot \frac{1}{5} = \frac{1}{\sqrt{n}}$ $\displaystyle \frac{3}{1.96}\cdot \frac{1}{5} = \frac{1}{\sqrt{n}}$

$n =\left ( \frac{3}{1.96 \cdot 5} \right )^{-2} = 10.671$ $\displaystyle n =\left ( \frac{3}{1.96 \cdot 5} \right )^{-2} = 10.671$

The minimum sample size is 10.671 $\displaystyle 10.671$ rounded up, which is $\displaystyle 11$ . If you are unsure on problems like these, you can check the margin of error for your answer rounded down and then rounded up (in this case, for n = 10 $\displaystyle n = 10$ and n = 11 $\displaystyle n = 11$ .)

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Example Question #61 : Ap Statistics

Jim calculated a $95\%$ $\displaystyle 95\%$ confidence interval for the mean height in inches of boys in his high school. He is not sure how to interpret this interval. Which of the following explains the meaning of $95\%$ $\displaystyle 95\%$ confidence.

Possible Answers:

More information is needed.

There is a $95\%$ $\displaystyle 95\%$ chance that Jim's interval contains the true mean height.

$95\%$ $\displaystyle 95\%$ of boys' heights fall with the interval Jim calculated.

$95\%$ $\displaystyle 95\%$ of all possible sample means fall within Jim's interval.

In the long run, $95\%$ $\displaystyle 95\%$ of all confidence intervals calculated from the same population will contain the true mean height.

Correct answer:

In the long run, $95\%$ $\displaystyle 95\%$ of all confidence intervals calculated from the same population will contain the true mean height.

Explanation:

95% confidence means that the methods Jim uses to calculate his confidence interval give him correct results 95% of the time. It does not mean that there is a 95% chance that the true mean will be inside the interval. It also does not mean that 95% of all heights or possible sample means fall within the interval.

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Example Question #2 : Confidence Intervals And Regression

The number of hamburgers served by McGregors per day is normally distributed and has a mean of 3,250 $\displaystyle 3,250$ hamburgers and a standard deviation of 320. $\displaystyle 320.$ Find the range of customers served on the middle $\displaystyle 50$ percent of days.

Possible Answers:

$\displaystyle (2,610, 3,890)$

$\displaystyle (2,930, 3,570)$

$\displaystyle (2,450, 4050)$

$\displaystyle (3,034, 3,466)$

$\displaystyle (2,290, 4,210)$

Correct answer:

$\displaystyle (3,034, 3,466)$

Explanation:

First, find the first quartile of the distribution.

$P(X < Q_{1}) = 0.25 \Rightarrow Q_{1} = \mu - Z_{0.25}\sigma = 3,250 - 0.675(320) = 3,034$ $\displaystyle P(X < Q_{1}) = 0.25 \Rightarrow Q_{1} = \mu - Z_{0.25}\sigma = 3,250 - 0.675(320) = 3,034$

Then, find the third quartile of the distribution.

$P(X < Q_{3}) = 0.75 \Rightarrow Q_{3} = \mu + Z_{0.25}\sigma = 3,250 + 0.675(320) = 3,466$ $\displaystyle P(X < Q_{3}) = 0.75 \Rightarrow Q_{3} = \mu + Z_{0.25}\sigma = 3,250 + 0.675(320) = 3,466$

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AP Statistics : Confidence Intervals and Regression

Study concepts, example questions & explanations for AP Statistics

All AP Statistics Resources

Example Questions

Example Question #1 : How To Find Confidence Intervals For The Slope Of A Least Squares Regression Line

Example Question #1 : How To Find Confidence Intervals

Example Question #12 : Confidence Intervals

Example Question #61 : Ap Statistics

Example Question #2 : Confidence Intervals And Regression

All AP Statistics Resources

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