How to find confidence intervals - AP Statistics

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Question

Which of the following statements are correct about confidence intervals?

Answer

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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Question

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?

Answer

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

From the problem, we can tell that $z^{*} = 1.96$ 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 .)

Compare your answer with the correct one above

Question

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.

Answer

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.

Compare your answer with the correct one above

Question

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.

Answer

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.

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

Compare your answer with the correct one above

Question

Which of the following statements are correct about confidence intervals?

Answer

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.

Compare your answer with the correct one above

Question

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?

Answer

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

From the problem, we can tell that $z^{*} = 1.96$ 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 .)

Compare your answer with the correct one above

Question

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.

Answer

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.

Compare your answer with the correct one above

Question

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.

Answer

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.

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

Compare your answer with the correct one above

Question

Which of the following statements are correct about confidence intervals?

Answer

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.

Compare your answer with the correct one above

Question

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?

Answer

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

From the problem, we can tell that $z^{*} = 1.96$ 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 .)

Compare your answer with the correct one above

Question

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.

Answer

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.

Compare your answer with the correct one above

Question

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.

Answer

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.

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

Compare your answer with the correct one above

Question

Which of the following statements are correct about confidence intervals?

Answer

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.

Compare your answer with the correct one above

Question

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?

Answer

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

From the problem, we can tell that $z^{*} = 1.96$ 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 .)

Compare your answer with the correct one above

Question

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.

Answer

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.

Compare your answer with the correct one above

Question

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.

Answer

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.

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

Compare your answer with the correct one above

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How to find confidence intervals - AP Statistics

Which of the following statements are correct about confidence intervals?

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?

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.

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.

First, find the first quartile of the distribution. Then, find the third quartile of the distribution.

Which of the following statements are correct about confidence intervals?

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?

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.

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.

First, find the first quartile of the distribution. Then, find the third quartile of the distribution.

Which of the following statements are correct about confidence intervals?

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?

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.

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.

First, find the first quartile of the distribution. Then, find the third quartile of the distribution.

Which of the following statements are correct about confidence intervals?

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?

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.

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.

First, find the first quartile of the distribution. Then, find the third quartile of the distribution.

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.

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

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.

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

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.

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

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.

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