B1.5
MARGIN OF ERROR FOR THE DIFFERENCE OF POPULATION MEANS
The question:
Data collected from a sample includes the weights (in pounds) of samples of regular coke and regular Pepsi. Use a .01 significance level to test the claim that the mean weight of regular coke is different from the mean of regular Pepsi given:
Regular Coke Pepsi
n 36 36
x .81682 .82410
s .007507 .005701
The question stem: What are they asking you to determine?
Use a .01 significance level to test the claim that the mean weight of regular coke is different from the mean of regular Pepsi.
What are you given?
Data collected from a sample includes the weights (in pounds) of samples of regular coke and regular Pepsi. Use a .01 significance level to test the claim that the mean weight of regular coke is different from the mean of regular Pepsi given:
Regular Coke Pepsi
n 36 36
x .81682 .82410
s .007507 .005701
*Independent samples
*Note: In these types of questions you must check the assumptions
The claim is that there is a difference between the weights of the cans which can be represented as
* Regular Coke Pepsi
n 36 36
x .81682 .82410
s .007507 .005701
*99% confidence in the results, which means critical value is 2.545
The formula: Two Sample ZInterval (2-SampZInterval)
Identify your variables:
Question type: 2-Samp ZInterval, Note: You can transform the equation to a z or t interval by using the appropriate statistical measure for large versus small samples.
Calculation: Using the T1-83: 2Samp TInterval
Select STAT -> TEST -> 2 Samp TInterval. Note select stats note data in this case You will need , calculate
This gives you a confidence interval of
Interpretation:
We are 90% confident that the true mean relating to player opinion falls between (-.55, .215)
MARGIN OF ERROR: IF YOU NOTICED ALREADY THE MARGIN OF ERROR IS MERELY THE SECOND PORTION OF THE CONFIDENCE INTERVAL FORMULA. SIMPLY ELIMINATE THE PLUS/MINUS THE STATISTIC FROM THE FRONT PORTION AND YOU ARE LEFT WITH THE MARGIN OR ERROR FORMULA
