Construct a confidence interval for $\mu _ { d }$ the mean of the differences d for the population of paired data. Assume that the population of paired differences is normally distributed. Using the sample paired data below, construct a 90% confidence interval for the population mean of all differences.
$\begin{array} { l | l l l l l }
\mathrm { A } & 5.0 & 5.1 & 4.6 & 3.5 & 6.0 \
\hline \mathrm { B } & 4.7 & 3.9 & 4.2 & 4.2 & 3.7
\end{array}$
A) $0.22

Question

Construct a confidence interval for $\mu _ { d }$ the mean of the differences d  for the population of paired data. Assume that the population of paired differences is normally distributed. Using the sample paired data below, construct a  90%  confidence interval for the population mean of all differences.
$\begin{array} { l | l l l l l } 
\mathrm { A } & 5.0 & 5.1 & 4.6 & 3.5 & 6.0 \
\hline \mathrm { B } & 4.7 & 3.9 & 4.2 & 4.2 & 3.7
\end{array}$
A) $0.22 < \mu _ { d } < 7.48$
B) $- 0.37 < \mu _ { d } < 1.77$
C) $- 0.07 < \mu _ { d } < 1.47$
D) $- 0.31 < \mu _ { d } < 1.71$

Quizplus · Accepted Answer

First, calculate the differences $d$ for each pair: $0.3, 1.2, 0.4, -0.7, 2.3$. The mean of these differences $\bar{d}$ is $0.7$, and the standard deviation $s_d$ is approximately $1.02$. For a 90% confidence interval and $n = 5$, the t-value from the t-distribution table (with $n-1 = 4$ degrees of freedom) is approximately $2.132$. The confidence interval is calculated using $\bar{d} \pm t \cdot \frac{s_d}{\sqrt{n}}$, which gives $-0.37 < \mu_d < 1.77$.

Construct a Confidence Interval For μd\mu _ { d }μd​ The Mean of the Differences D for the Population

Construct a Confidence Interval For $\mu _ { d }$ The Mean of the Differences D for the Population