Construct a 95% confidence interva

Statistics
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A researcher was interested in comparing the GPAs of students at two different colleges. Independent simple random samples of 8 students from college A and 13 students from college B yielded the following GPAs.

College ACollege B
3.73.82.8
3.23.24.0
3.03.03.6
2.53.92.6
2.73.84.0
3.62.53.6
2.83.9
3.4


    
Construct a 95% confidence interval for the difference between the mean GPA of college A students and the mean GPA of college B students.

Jul 29th, 2015

Thank you for the opportunity to help you with your question!

To calculate a confidence interval for the population mean (average), under these conditions, do the following: Determine the confidence level and degrees of freedom and then find the appropriate t*-value. Find the sample mean. Multiply t* times s and divide that by the square root of n

You estimate the population mean,

by using a sample mean,

plus or minus a margin of error. The result is called a confidence interval for the population mean,

In many situations, you don’t know

so you estimate it with the sample standard deviation, s; and/or the sample size is small (less than 30), and you can’t be sure your data came from a normal distribution. (In the latter case, the Central Limit Theorem can’t be used.) In either situation, you can’t use a z*-value from the standard normal (Z-) distribution as your critical value anymore; you have to use a larger critical value than that, because of not knowing what

is and/or having less data.

The formula for a confidence interval for one population mean in this case 

is the critical t*-value from the t-distribution with n – 1 degrees of freedom (where n is the sample size).


Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 29th, 2015

Is this the correct answer? 

-0.81 < μ1 - μ2 < 0.15

Jul 29th, 2015

exactly that, you must have used the formula correctlt

Jul 29th, 2015

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