Determine whether the samples are independent or dependent.
-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. $$\begin{array} { | c | c c | }
\hline \text { College A } & { \text { College B } } \
\hline 3.7 & 3.8 & 2.8 \
3.2 & 3.2 & 4.0 \
3.0 & 3.0 & 3.6 \
2.5 & 3.9 & 2.6 \
2.7 & 3.8 & 4.0 \
3.6 & 2.5 & 3.6 \
2.8 & 3.9 & \
3.4 & & \
\hline
\end{array}$$ Construct a 95% confidence interval for $$\mu _ { 1 } - \mu _ { 2 }$$ , the difference between the mean GPA of college A students
And the mean GPA of college B students. (Note: $\bar { x } _ { 1 } = 3.1125 , \bar { x } _ { 2 } = 3.4385 , \mathrm {~s} _ { 1 } = 0.4357$, $\mathrm { s } _ { 2 } = 0.5485$.)
A) $- 0.75

Question

Determine whether the samples are independent or dependent.
-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. $$\begin{array} { | c | c c | } 
\hline \text { College A } & { \text { College B } } \
\hline 3.7 & 3.8 & 2.8 \
3.2 & 3.2 & 4.0 \
3.0 & 3.0 & 3.6 \
2.5 & 3.9 & 2.6 \
2.7 & 3.8 & 4.0 \
3.6 & 2.5 & 3.6 \
2.8 & 3.9 & \
3.4 & & \
\hline
\end{array}$$ Construct a 95% confidence interval for $$\mu _ { 1 } - \mu _ { 2 }$$ , the difference between the mean GPA of college A students
And the mean GPA of college B students. (Note: $\bar { x } _ { 1 } = 3.1125 , \bar { x } _ { 2 } = 3.4385 , \mathrm {~s} _ { 1 } = 0.4357$, $\mathrm { s } _ { 2 } = 0.5485$.)
A) $- 0.75 < \mu _ { 1 } - \mu _ { 2 } < 0.10$
B) $- 0.78 < \mu _ { 1 } - \mu _ { 2 } < 0.13$
C) $- 0.81 < \mu _ { 1 } - \mu _ { 2 } < 0.15$
D) $- 0.70 < \mu _ { 1 } - \mu _ { 2 } < 0.05$

Quizplus · Accepted Answer

The samples are independent because they are from two different colleges. To construct the confidence interval, we will use the formula $$(\bar{x_1}-\bar{x_2}) \pm t_{df}^{*} \sqrt{ \frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}.$$ The degrees of freedom are $$df = \frac{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2})^2}{\frac{(s_1^2/n_1)^2}{n_1-1}+\frac{(s_2^2/n_2)^2}{n_2-1}}\approx 17.6.$$ The critical value with 17.6 degrees of freedom and a 95% confidence level is approximately 2.1098. Substituting the given values, we get $$(3.1125-3.4385) \pm 2.1098 \sqrt{ \frac{(0.4357)^2}{8}+\frac{(0.5485)^2}{13}}.$$ Simplifying, we get $$-0.78 < \mu _ { 1 } - \mu _ { 2 } < 0.13.$$ Therefore, the answer is B.

Determine Whether the Samples Are Independent or Dependent Construct a 95% Confidence Interval For