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 A  College B  
3.7  3.8  2.8 
3.2  3.2  4.0 
3.0  3.0  3.6 
2.5  3.9  2.6 
2.7  3.8  4.0 
3.6  2.5  3.6 
2.8  3.9  
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.
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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 tdistribution 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.
Is this the correct answer?
0.81 < μ1  μ2 < 0.15 
exactly that, you must have used the formula correctlt
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