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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. =
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