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

To construct a 90% confidence interval for the population mean of all differences, we first calculate the differences between each pair of A and B, then find the mean ($\bar{d}$) and standard deviation (s) of these differences, and finally use the t-distribution because the population standard deviation is unknown and the sample size is small.Differences (d) = A - B = {0.3, 1.2, 0.4, -0.7, 2.3}Mean of differences ($\bar{d}$) = (0.3 + 1.2 + 0.4 - 0.7 + 2.3) / 5 = 3.5 / 5 = 0.7Standard deviation (s) of differences = sqrt([(0.3-0.7)^2 + (1.2-0.7)^2 + (0.4-0.7)^2 + (-0.7-0.7)^2 + (2.3-0.7)^2] / (5-1))= sqrt([0.16 + 0.25 + 0.09 + 1.96 + 2.56] / 4)= sqrt(4.02 / 4)= sqrt(1.005)= 1.0025Standard error (SE) = s / sqrt(n) = 1.0025 / sqrt(5) = 1.0025 / 2.236 = 0.4485For a 90% confidence interval and df = n - 1 = 5 - 1 = 4, the t-value from the t-distribution table is approximately 2.132.Margin of error (ME) = t * SE = 2.132 * 0.4485 ≈ 0.956Therefore, the 90% confidence interval for $\mu_d$ is $\bar{d} \pm ME = 0.7 \pm 0.956$, which gives us:Lower limit = 0.7 - 0.956 = -0.256Upper limit = 0.7 + 0.956 = 1.656Rounding to two decimal places, the confidence interval is approximately $-0.26 < \mu_d < 1.66$, which is closest to option B) $- 0.37 < \mu _ { d } < 1.77$.

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

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