Confidence Intervals - AP Statistics

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Question

Answer

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Question

The confidence interval created for the difference in means between two training programs for middle distance college runners is $\small \left ( -4.5, 6.5\right )$ . The variable being measured is the improvement in seconds of mile times over the course of a season. One program has more speed work and intervals, while the other focuses more on distance training.

What does the confidence interval tell us about the difference in the two programs?

Answer

For there to be a statistically significant difference in the training programs, the 95% confidence interval cannot include zero. $\small \left ( -4.5, 6.5\right )$ includes zero, so we can't say that one program is significantly better than the other.

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Question

Suppose you have a normally distributed variable with known variance. How many standard errors do you need to add and subtract from the sample mean so that you obtain 95% confidence intervals?

Answer

To obtain 95% confidence intervals for a normal distribution with known variance, you take the mean and add/subtract $1.96\times standard\ error$ . This is because 95% of the values drawn from a normally distributed sampling distribution lie within 1.96 standard errors from the sample mean.

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Question

An automotive engineer wants to estimate the cost of repairing a car that experiences a 25 MPH head-on collision. He crashes 24 cars, and the average repair is $11,000. The standard deviation of the 24-car sample is $2,500.

Provide a 98% confidence interval for the true mean cost of repair.

Answer

Standard deviation for the samle mean:

$2500/\sqrt{24}=510.2$

Since n < 30, we must use the t-table (not the z-table).

The 98% t-value for n=24 is 2.5.

$11,000\pm2.5\times510.2=11,000\pm1275.5$

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Question

The population standard deviation is 7. Our sample size is 36.

What is the 95% margin of error for:

the population mean

the sample mean

Answer

For 95% confidence, Z = 1.96.

The population M.O.E. =

$1.96 \times 7 = 13.720$

The sample standard deviation =

$7\div\sqrt{36}=1.167$

The sample M.O.E. =

$1.96\times1.167=2.287$

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Question

300 hundred eggs were randomly chosen from a gravid female salmon and individually weighed. The mean weight was 0.978 g with a standard deviation of 0.042. Find the 95% confidence interval for the mean weight of the salmon eggs (because it is a large n, use the standard normal distribution).

Answer

Because we have such a large sample size, we are using the standard normal or z-distribution to calculate the confidence interval.

Formula:

$\bar{x} \pm z \cdot \frac{\sigma}{\sqrt{n}}$

$\bar{x}= 0.978 g$

$\sigma= 0.042$

We must find the appropriate z-value based on the given $\alpha$ for 95% confidence:

$1-\frac{\alpha}{2}= 1-\left(\frac{0.05}{2}\right)= 0.975$

Then, find the associated z-score using the z-table for

$z _{0.975} = 1.96$

Now we fill in the formula with our values from the problem to find the 95% CI.

$0.978 \pm 1.96 \left( \frac{0.042}{\sqrt{300}}\right)= (0.973 g, 0.9827 g)$

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Question

A sample of observations of 02 consumption by adult western fence lizards gave the following statistics:

Find the confidence limit for the mean 02 consumption by adult western fence lizards.

Answer

Because we are only given the sample standard deviation we will use the t-distribution to calculate the confidence interval.

Appropriate Formula:

$= \bar{x} \pm t \cdot \frac{s}{\sqrt{n}}$

Now we must identify our variables:

$\bar{x}=249$

We must find the appropriate t-value based on the given $\alpha$

t-value at 90% confidence:

$t_{\frac{\alpha }{2}, n-1}= t _{\frac{0.10}{2}}= t_{0.05, 55}$

Look up t-value for 0.05, 55 , so t-value= ~ 1.6735

90% CI becomes:

$249 \pm 1.6735\left(\frac{24}{\sqrt{56}}\right)$

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Question

Subject Horn Length (in) Subject Horn Length (in)
1 19.1 11 11.6
2 14.7 12 18.5
3 10.2 13 28.7
4 16.1 14 15.3
5 13.9 15 13.5
6 12.0 16 7.7
7 20.7 17 17.2
8 8.6 18 19.0
9 24.2 19 20.9
10 17.3 20 21.3

The data above represents measurements of the horn lengths of African water buffalo that were raised on calcium supplements. Construct a 95% confidence interval for the population mean for horn length after supplments.

Subject	Horn Length (in)	Subject	Horn Length (in)
1	19.1	11	11.6
2	14.7	12	18.5
3	10.2	13	28.7
4	16.1	14	15.3
5	13.9	15	13.5
6	12.0	16	7.7
7	20.7	17	17.2
8	8.6	18	19.0
9	24.2	19	20.9
10	17.3	20	21.3

Answer

First you must calculate the sample mean and sample standard deviation of the sample.

$\bar{x}=16.53$

$\sigma=5.29$

Because we do not know the population standard deviation we will use the t-distribution to calculate the confidence intervals. We must use standard error in this formula because we are working with the standard deviation of the sampling distribution.

Formula:

$\bar{x} \pm t\cdot \frac{s}{\sqrt{n}}$

To find the appropriate t-value for 95% confidence interval:

$\alpha =0.05, \frac{0.05}{2}=0.025$

Look up $t_{\frac{\alpha }{2}, 19}$ in t-table and the corresponding t-value = 2.093.

Thus the 95% confidence interval is:

$16.53 \pm 2.093\left(\frac{5.29}{\sqrt{19}}\right)= (13.99 mm Hg, 19.07 mm Hg)$

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Question

In a simple random sample of people, are left-handed. Find a confidence interval for the true proportion of left-handed people in the entire population.

Answer

Step 1: Determine the approporiate formula. In this case, the statistic is a proportion because the question says 261 people out of 1000 are left-handed. We want to use a confidence interval for proportions. The equation is

$\hat{p}\pm z^{} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Step 2: $\hat{p}$ , or the sample proportion is equal to $\frac{261}{1000} = 0.261$

Step 3: The questions asks for a 95% confidence interval. You can assume a normal distribution because of the large sample size. Therefore, in a normal distribution, 95% of the data is contained within $z=\pm 1.96$ .

$z^{}=1.96$

Step 4: Substitue all the values into the equation to get the answer.

$0.261\pm1.96 \sqrt{\frac{(0.261)(0.739)}{1000}}$

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Question

A health insurance executive suspects that 50% of insurance applications are incomplete (i.e. critical data is missing from the application). He wants to gather a sample to estimate the true proportion. He wants to be within 5% of the true mean with 95% confidence.

What sample size does he need?

Answer

$n=pq\times(Z/E)^{2}$

$n=.5\times.5\times(1.96/.05)^{2} = 384.16 = 385$

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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 .)

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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.

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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.

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

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Question

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 $\beta$ .

Answer

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 $\beta$ .

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Question

Answer

No explanation available

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Question

The confidence interval created for the difference in means between two training programs for middle distance college runners is $\small \left ( -4.5, 6.5\right )$ . The variable being measured is the improvement in seconds of mile times over the course of a season. One program has more speed work and intervals, while the other focuses more on distance training.

What does the confidence interval tell us about the difference in the two programs?

Answer

For there to be a statistically significant difference in the training programs, the 95% confidence interval cannot include zero. $\small \left ( -4.5, 6.5\right )$ includes zero, so we can't say that one program is significantly better than the other.

Compare your answer with the correct one above

Question

Suppose you have a normally distributed variable with known variance. How many standard errors do you need to add and subtract from the sample mean so that you obtain 95% confidence intervals?

Answer

To obtain 95% confidence intervals for a normal distribution with known variance, you take the mean and add/subtract $1.96\times standard\ error$ . This is because 95% of the values drawn from a normally distributed sampling distribution lie within 1.96 standard errors from the sample mean.

Compare your answer with the correct one above

Question

An automotive engineer wants to estimate the cost of repairing a car that experiences a 25 MPH head-on collision. He crashes 24 cars, and the average repair is $11,000. The standard deviation of the 24-car sample is $2,500.

Provide a 98% confidence interval for the true mean cost of repair.

Answer

Standard deviation for the samle mean:

$2500/\sqrt{24}=510.2$

Since n < 30, we must use the t-table (not the z-table).

The 98% t-value for n=24 is 2.5.

$11,000\pm2.5\times510.2=11,000\pm1275.5$

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Confidence Intervals - AP Statistics

No explanation available

For there to be a statistically significant difference in the training programs, the 95% confidence interval cannot include zero. includes zero, so we can't say that one program is significantly better than the other.

Suppose you have a normally distributed variable with known variance. How many standard errors do you need to add and subtract from the sample mean so that you obtain 95% confidence intervals?

To obtain 95% confidence intervals for a normal distribution with known variance, you take the mean and add/subtract . This is because 95% of the values drawn from a normally distributed sampling distribution lie within 1.96 standard errors from the sample mean.

An automotive engineer wants to estimate the cost of repairing a car that experiences a 25 MPH head-on collision. He crashes 24 cars, and the average repair is $11,000. The standard deviation of the 24-car sample is $2,500. Provide a 98% confidence interval for the true mean cost of repair.

Standard deviation for the samle mean: Since n < 30, we must use the t-table (not the z-table). The 98% t-value for n=24 is 2.5.

The population standard deviation is 7. Our sample size is 36. What is the 95% margin of error for: the population mean the sample mean

For 95% confidence, Z = 1.96. The population M.O.E. = The sample standard deviation = The sample M.O.E. =

300 hundred eggs were randomly chosen from a gravid female salmon and individually weighed. The mean weight was 0.978 g with a standard deviation of 0.042. Find the 95% confidence interval for the mean weight of the salmon eggs (because it is a large n, use the standard normal distribution).

A sample of observations of 02 consumption by adult western fence lizards gave the following statistics: Find the confidence limit for the mean 02 consumption by adult western fence lizards.

No explanation available

In a simple random sample of people, are left-handed. Find a confidence interval for the true proportion of left-handed people in the entire population.

A health insurance executive suspects that 50% of insurance applications are incomplete (i.e. critical data is missing from the application). He wants to gather a sample to estimate the true proportion. He wants to be within 5% of the true mean with 95% confidence. What sample size does he need?

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.

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

To construct 95% confidence intervals for , we simply take the coefficient and add/subtract . This is because 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 .

No explanation available

For there to be a statistically significant difference in the training programs, the 95% confidence interval cannot include zero. includes zero, so we can't say that one program is significantly better than the other.

Suppose you have a normally distributed variable with known variance. How many standard errors do you need to add and subtract from the sample mean so that you obtain 95% confidence intervals?

To obtain 95% confidence intervals for a normal distribution with known variance, you take the mean and add/subtract . This is because 95% of the values drawn from a normally distributed sampling distribution lie within 1.96 standard errors from the sample mean.

An automotive engineer wants to estimate the cost of repairing a car that experiences a 25 MPH head-on collision. He crashes 24 cars, and the average repair is $11,000. The standard deviation of the 24-car sample is $2,500. Provide a 98% confidence interval for the true mean cost of repair.

Standard deviation for the samle mean: Since n < 30, we must use the t-table (not the z-table). The 98% t-value for n=24 is 2.5.

For there to be a statistically significant difference in the training programs, the 95% confidence interval cannot include zero. $\small \left ( -4.5, 6.5\right )$ includes zero, so we can't say that one program is significantly better than the other.

To obtain 95% confidence intervals for a normal distribution with known variance, you take the mean and add/subtract $1.96\times standard\ error$ . This is because 95% of the values drawn from a normally distributed sampling distribution lie within 1.96 standard errors from the sample mean.

An automotive engineer wants to estimate the cost of repairing a car that experiences a 25 MPH head-on collision. He crashes 24 cars, and the average repair is $11,000. The standard deviation of the 24-car sample is $2,500.

Provide a 98% confidence interval for the true mean cost of repair.

Standard deviation for the samle mean:

$2500/\sqrt{24}=510.2$

Since n < 30, we must use the t-table (not the z-table).

The 98% t-value for n=24 is 2.5.

$11,000\pm2.5\times510.2=11,000\pm1275.5$

The population standard deviation is 7. Our sample size is 36.

What is the 95% margin of error for:

the population mean

the sample mean

For 95% confidence, Z = 1.96.

The population M.O.E. =

$1.96 \times 7 = 13.720$

The sample standard deviation =

$7\div\sqrt{36}=1.167$

The sample M.O.E. =

$1.96\times1.167=2.287$

A sample of observations of 02 consumption by adult western fence lizards gave the following statistics:

Find the confidence limit for the mean 02 consumption by adult western fence lizards.

A health insurance executive suspects that 50% of insurance applications are incomplete (i.e. critical data is missing from the application). He wants to gather a sample to estimate the true proportion. He wants to be within 5% of the true mean with 95% confidence.

What sample size does he need?

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 $\beta$ .

For there to be a statistically significant difference in the training programs, the 95% confidence interval cannot include zero. $\small \left ( -4.5, 6.5\right )$ includes zero, so we can't say that one program is significantly better than the other.

To obtain 95% confidence intervals for a normal distribution with known variance, you take the mean and add/subtract $1.96\times standard\ error$ . This is because 95% of the values drawn from a normally distributed sampling distribution lie within 1.96 standard errors from the sample mean.

An automotive engineer wants to estimate the cost of repairing a car that experiences a 25 MPH head-on collision. He crashes 24 cars, and the average repair is $11,000. The standard deviation of the 24-car sample is $2,500.

Provide a 98% confidence interval for the true mean cost of repair.

Standard deviation for the samle mean:

$2500/\sqrt{24}=510.2$

Since n < 30, we must use the t-table (not the z-table).

The 98% t-value for n=24 is 2.5.

$11,000\pm2.5\times510.2=11,000\pm1275.5$