Confidence Intervals and Mean - AP Statistics

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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?

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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.

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

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No explanation available

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?

Tap to see back →

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.

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.

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

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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).

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.

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

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No explanation available

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?

Tap to see back →

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.

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.

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

Confidence Intervals and Mean - 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

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

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.

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.

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.

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