AP Statistics : How to find confidence intervals for a mean

Study concepts, example questions & explanations for AP Statistics

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Example Questions

Example Question #1 : Confidence Intervals And Mean

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?

Possible Answers:

Correct answer:

Explanation:

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.

Possible Answers:

 

Correct answer:

Explanation:

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 #1 : 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).

 

Possible Answers:

Correct answer:

Explanation:

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 #1 : Confidence Intervals

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.

Possible Answers:

Correct answer:

Explanation:

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 #1 : 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.

Possible Answers:

Correct answer:

Explanation:

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

Possible Answers:

1) 11

2) 3

1) 12.266

2) 3.711

1) 14.567

2) 4.445

1) 15.554

2) 3.656

1) 13.720

2) 2.287

Correct answer:

1) 13.720

2) 2.287

Explanation:

For 95% confidence, Z = 1.96.

1) The population M.O.E. =

 

2) The sample standard deviation = 

The sample M.O.E. = 

 

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