Estimation - AP Statistics
Card 0 of 88
The next local election is predicted to have a 
split in votes for the top two candidates. How many people should be polled to obtain a margin of error of 3% at the 95% confidence level?
The next local election is predicted to have a
If you remember the required formula, this problem is rather simple. Plug in the given numbers and simplify:
If you don't remember the formula, this problem is more challenging. Your best bet in this case is to construct the confidence interval and rearrange to solve for the required sample size.
If you remember the required formula, this problem is rather simple. Plug in the given numbers and simplify:
If you don't remember the formula, this problem is more challenging. Your best bet in this case is to construct the confidence interval and rearrange to solve for the required sample size.
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When dealing with confidence intervals, the margin of error gets smaller when z* gets ________ and n gets _________.
When dealing with confidence intervals, the margin of error gets smaller when z* gets ________ and n gets _________.
For confidence intervals, a small margin of error is preferred, as it indicates that the parameter of interest has been narrowed down to a precise interval. Having a large sample population to work from as well as a small z* (z* gets smaller as the confidence level percent gets lower) can help obtain a small margin of error.
For confidence intervals, a small margin of error is preferred, as it indicates that the parameter of interest has been narrowed down to a precise interval. Having a large sample population to work from as well as a small z* (z* gets smaller as the confidence level percent gets lower) can help obtain a small margin of error.
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What is 
, or the expected value of 
for any distribution?
What is
is 
, or the mean, of the population. This makes sense since 
literally means the expected value of 
. The mean is the expected value of 
.
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The 
confidence interval created for the difference in means between two training programs for middle distance college runners is 
. 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?
The 
What does the confidence interval tell us about the difference in the two programs?
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.
For there to be a statistically significant difference in the training programs, the 95% confidence interval cannot include zero.
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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?
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.
To obtain 95% confidence intervals for a normal distribution with known variance, you take the mean and add/subtract
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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.
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.
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.
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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
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. =
For 95% confidence, Z = 1.96.
- The population M.O.E. =
- The sample standard deviation =
The sample M.O.E. =
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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).
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).
Because we have such a large sample size, we are using the standard normal or z-distribution to calculate the confidence interval.
Formula:
We must find the appropriate z-value based on the given 
for 95% confidence:
Then, find the associated z-score using the z-table for
Now we fill in the formula with our values from the problem to find the 95% CI.
Because we have such a large sample size, we are using the standard normal or z-distribution to calculate the confidence interval.
Formula:
We must find the appropriate z-value based on the given
Then, find the associated z-score using the z-table for
Now we fill in the formula with our values from the problem to find the 95% CI.
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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 sample of
Find the
Because we are only given the sample standard deviation we will use the t-distribution to calculate the confidence interval.
Appropriate Formula:
Now we must identify our variables:
We must find the appropriate t-value based on the given
t-value at 90% confidence:
Look up t-value for 0.05, 55 , so t-value= ~ 1.6735
90% CI becomes:
Because we are only given the sample standard deviation we will use the t-distribution to calculate the confidence interval.
Appropriate Formula:
Now we must identify our variables:
We must find the appropriate t-value based on the given
t-value at 90% confidence:
Look up t-value for 0.05, 55 , so t-value= ~ 1.6735
90% CI becomes:
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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 |
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.
First you must calculate the sample mean and sample standard deviation of the sample.
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:
To find the appropriate t-value for 95% confidence interval:
Look up 
in t-table and the corresponding t-value = 2.093.
Thus the 95% confidence interval is:
First you must calculate the sample mean and sample standard deviation of the sample.
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:
To find the appropriate t-value for 95% confidence interval:
Look up
Thus the 95% confidence interval is:
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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.
In a simple random sample of
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
Step 2: 
, or the sample proportion is equal to 
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 
.
Step 4: Substitue all the values into the equation to get the 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
Step 2:
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
Step 4: Substitue all the values into the equation to get the answer.
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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?
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?
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Which of the following statements are correct about confidence intervals?
Which of the following statements are correct about confidence intervals?
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.
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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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?
You are asked to create a
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 
. We can then solve for 
algebraically:
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 
.)
Keep in mind that the margin of error for a confidence interval based on a normal population is equal to
From the problem, we can tell that
The minimum sample size is
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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.
Jim calculated a
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.
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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