How to find confidence intervals for a mean - AP Statistics

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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?

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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.

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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

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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. =

$1.96\times1.167=2.287$

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).

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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)$

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.

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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:

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}$}\right)$

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

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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 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}$}\right)= (13.99 mm Hg, 19.07 mm Hg)$

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?

Tap to see back →

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.

Tap to see back →

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

Tap to see back →

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. =

$1.96\times1.167=2.287$

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).

Tap to see back →

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)$

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.

Tap to see back →

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:

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}$}\right)$

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

Tap to see back →

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 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}$}\right)= (13.99 mm Hg, 19.07 mm Hg)$

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?

Tap to see back →

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.

Tap to see back →

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

Tap to see back →

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. =

$1.96\times1.167=2.287$

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).

Tap to see back →

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)$

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.

Tap to see back →

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:

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}$}\right)$

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

Tap to see back →

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 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}$}\right)= (13.99 mm Hg, 19.07 mm Hg)$

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?

Tap to see back →

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.

Tap to see back →

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.

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

How to find confidence intervals for a mean - AP Statistics

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.

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.

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.

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.

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.

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.

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.

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.

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