We want to know if college students who are Facebook users have different GPA\'s

Question

We want to know if college students who are Facebook users have different GPA's from college students who are not Facebook users. A random sample of 41 College Facebook users resulted in a mean GPA of  x = 3.06. A random sample of 35 non-Facebook users resulted in a mean GPA of y =3.37. It is recognized that the standard deviation of Facebook user GPA is x = 0.55 while it is recognized that the standard deviation of non-Facebook user GPA is y = 0.21 . The true (unknown) mean Facebook user GPA is x while the true (unknown) non-facebook user GPA is y.

a)Calculate the variance of the random variable X which is the mean of the 41 Facebook user GPA's.  

b)Calculate the variance of the random variable Y, which is the mean of the 35 non-Facebook user GPA's.  

c) Calculate the variance of X - Y?  

d) Calculate the standard deviation of X - Y?  

e) If we wish to create a 94% confidence interval for x - y then what is the z critical value used?

f)Create a 94% confidence interval for x - y .

g) What is the length of your 94% confidence interval for x - y?

h) If we used this data to test H0: x - y =0 against the alternative Ha: x - y< 0 then what would the value of the calculated test statistic z have been?

i) If we used this data to test H0:x - y =0 against the alternative Ha: x - y<0 then what would the p value have been?

j)If we used this data to test H0: x - y =0 against the alternative Ha: x - y 0 then what would the p value have been?

k) Copy your R script for the above into the text box here.

Explanation / Answer

a) V(Xbar) V(X)/n [sd(x)]^2 /n =0.55^2/41 0.007378049 b) V(ybar) V(Y)/n [SD(Y)]^2/n =0.21^2/35 0.00126 c) V(X-Y) V(x) - V(Y) =0.007378-0.00126 0.006118 d) SD(X-Y) sqrt[V(X-Y)] =SQRT(0.006118) 0.078217645

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