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Example Questions
Example Question #1 : 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.
Example Question #1 : Confidence Intervals
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.
Example Question #3 : Confidence Intervals
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.
Example Question #4 : Confidence Intervals
A sample of observations of 0_{2} consumption by adult western fence lizards gave the following statistics:
Find the confidence limit for the mean 0_{2} consumption by adult western fence lizards.
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:
Example Question #5 : Confidence Intervals
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:
Example Question #1 : Confidence Intervals
The population standard deviation is 7. Our sample size is 36.
What is the 95% margin of error for:
1) the population mean
2) the sample mean
1) 14.567
2) 4.445
1) 12.266
2) 3.711
1) 15.554
2) 3.656
1) 11
2) 3
1) 13.720
2) 2.287
1) 13.720
2) 2.287
For 95% confidence, Z = 1.96.
1) The population M.O.E. =
2) The sample standard deviation =
The sample M.O.E. =
Example Question #7 : Confidence Intervals
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?
Zero is in the interval, so do not reject the null. There is no evidence that one program is better than the other.
is greater than , so reject the null. This is evidence that one program is better at reducing mile times than the other.
The mean improvement of second is too small to matter, so reject the null. There is no evidence that one program is better at reducing miles times.
Zero is not in the interval, so reject the null. This is evidence that one program is significantly better at reducing mile times.
The confidence interval is large, so one program is clearly better at reducing mile times than the other.
Zero is in the interval, so do not reject the null. There is no evidence that one program is better than the other.
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.
Example Question #8 : Confidence Intervals
Example Question #9 : Confidence Intervals
Example Question #10 : Confidence Intervals
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.
The sample is not sufficiently large. A confidence interval cannot be used.
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.
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