Nine students took the SAT. Their scores are listed below. Later on, they read a book on test preparation andretook the SAT. Their new scores are listed below. Construct a 95% confidence interval for μd. Assume that thedistribution is normally distributed. $$\begin{array} { l | c c c c c c c c c }
\text { Student } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \
\hline \text { Scores before reading book } & 720 & 860 & 850 & 880 & 860 & 710 & 850 & 1200 & 950 \
\hline \text { Scores after reading book } & 740 & 860 & 840 & 920 & 890 & 720 & 840 & 1240 & 970
\end{array}$$
A) (-30.503, -0.617)
B) (-20.341, 4.852)
C) (-10.321, 15.436)
D) (1.651, 30.590)

Question

Nine students took the SAT. Their scores are listed below. Later on, they read a book on test preparation andretook the SAT. Their new scores are listed below. Construct a 95% confidence interval for μd. Assume that thedistribution is normally distributed. $$\begin{array} { l | c c c c c c c c c } 
\text { Student } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \
\hline \text { Scores before reading book } & 720 & 860 & 850 & 880 & 860 & 710 & 850 & 1200 & 950 \
\hline \text { Scores after reading book } & 740 & 860 & 840 & 920 & 890 & 720 & 840 & 1240 & 970
\end{array}$$
A) (-30.503, -0.617)
B) (-20.341, 4.852)
C) (-10.321, 15.436)
D) (1.651, 30.590)

Quizplus · Accepted Answer

The correct answer is A because the 95% confidence interval for μd is (-30.503, -0.617). This interval is calculated using the formula:CI = (x̄d - z * (s/√n), x̄d + z * (s/√n))where x̄d is the sample mean of the differences, s is the sample standard deviation of the differences, n is the sample size, and z is the z-score corresponding to the desired confidence level.In this case, x̄d = -15.556, s = 14.944, n = 9, and z = 1.96. Plugging these values into the formula, we get:CI = (-15.556 - 1.96 * (14.944/√9), -15.556 + 1.96 * (14.944/√9)) = (-30.503, -0.617)

Nine Students Took the SAT A) (-30503, -0