Confidence Intervals and Regression - AP Statistics

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Which of the following statements are correct about confidence intervals?

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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.

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

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Keep in mind that the margin of error for a confidence interval based on a normal population is equal to , where is the -score corresponding to the desired confidence level.

From the problem, we can tell that and $\sigma = 5$ . We can then solve for algebraically:

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

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

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

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

The minimum sample size is rounded up, which is . 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 and .)

Jim calculated a 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 confidence.

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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.

The number of hamburgers served by McGregors per day is normally distributed and has a mean of hamburgers and a standard deviation of Find the range of customers served on the middle percent of days.

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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$

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$

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

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To construct 95% confidence intervals for $\beta$ , we simply take the coefficient and add/subtract $1.96\times the\ standard\ error\ of\ \beta$ . This is because $\beta$ 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 $\beta$ .

Which of the following statements are correct about confidence intervals?

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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.

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

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Keep in mind that the margin of error for a confidence interval based on a normal population is equal to , where is the -score corresponding to the desired confidence level.

From the problem, we can tell that and $\sigma = 5$ . We can then solve for algebraically:

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

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

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

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

The minimum sample size is rounded up, which is . 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 and .)

Jim calculated a 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 confidence.

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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.

The number of hamburgers served by McGregors per day is normally distributed and has a mean of hamburgers and a standard deviation of Find the range of customers served on the middle percent of days.

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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$

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$

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

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To construct 95% confidence intervals for $\beta$ , we simply take the coefficient and add/subtract $1.96\times the\ standard\ error\ of\ \beta$ . This is because $\beta$ 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 $\beta$ .

Which of the following statements are correct about confidence intervals?

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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.

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

Tap to see back →

Keep in mind that the margin of error for a confidence interval based on a normal population is equal to , where is the -score corresponding to the desired confidence level.

From the problem, we can tell that and $\sigma = 5$ . We can then solve for algebraically:

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

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

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

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

The minimum sample size is rounded up, which is . 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 and .)

Jim calculated a 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 confidence.

Tap to see back →

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.

The number of hamburgers served by McGregors per day is normally distributed and has a mean of hamburgers and a standard deviation of Find the range of customers served on the middle percent of days.

Tap to see back →

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$

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$

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

Tap to see back →

To construct 95% confidence intervals for $\beta$ , we simply take the coefficient and add/subtract $1.96\times the\ standard\ error\ of\ \beta$ . This is because $\beta$ 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 $\beta$ .

Which of the following statements are correct about confidence intervals?

Tap to see back →

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.

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

Tap to see back →

Keep in mind that the margin of error for a confidence interval based on a normal population is equal to , where is the -score corresponding to the desired confidence level.

From the problem, we can tell that and $\sigma = 5$ . We can then solve for algebraically:

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

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

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

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

The minimum sample size is rounded up, which is . 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 and .)

Jim calculated a 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 confidence.

Tap to see back →

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.

The number of hamburgers served by McGregors per day is normally distributed and has a mean of hamburgers and a standard deviation of Find the range of customers served on the middle percent of days.

Tap to see back →

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$

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$

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

Tap to see back →

To construct 95% confidence intervals for $\beta$ , we simply take the coefficient and add/subtract $1.96\times the\ standard\ error\ of\ \beta$ . This is because $\beta$ 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 $\beta$ .