Construct a 95% confidence interval for data sets A and B. Data sets A and B are dependent. $$\begin{array} { c | l l l l l }
\text { A } & 30 & 28 & 47 & 43 & 31 \
\hline \text { B } & 28 & 24 & 25 & 35 & 22
\end{array}$$ Assume that the paired data came from a population that is normally distributed.
A) (-0.696, 18.700)
B) (-1.324, 8.981)
C) (-0.113, 12.761)
D) (-15.341, 15.431)

Question

Construct a 95% confidence interval for data sets A and B. Data sets A and B are dependent. $$\begin{array} { c | l l l l l } 
\text { A } & 30 & 28 & 47 & 43 & 31 \
\hline \text { B } & 28 & 24 & 25 & 35 & 22
\end{array}$$ Assume that the paired data came from a population that is normally distributed.
A) (-0.696, 18.700)
B) (-1.324, 8.981)
C) (-0.113, 12.761)
D) (-15.341, 15.431)

Quizplus · Accepted Answer

To construct a 95% confidence interval for dependent data sets A and B, we first calculate the differences between each pair of observations, then find the mean and standard deviation of these differences, and finally use the t-distribution to calculate the interval because the sample size is small and the population standard deviation is unknown.Differences (A-B): 2, 4, 22, 8, 9Mean difference ($\bar{d}$): $ \frac{2 + 4 + 22 + 8 + 9}{5} = 9 $Standard deviation of differences (s): $ \sqrt{\frac{(2-9)^2 + (4-9)^2 + (22-9)^2 + (8-9)^2 + (9-9)^2}{4}} = \sqrt{\frac{49 + 25 + 169 + 1 + 0}{4}} = \sqrt{\frac{244}{4}} = \sqrt{61} \approx 7.810 $Standard error of the mean difference (SE): $ \frac{s}{\sqrt{n}} = \frac{7.810}{\sqrt{5}} \approx 3.492 $For a 95% confidence interval with 4 degrees of freedom (n-1 = 5-1), the t-value from the t-distribution table is approximately 2.776.Margin of error (ME): $ t \times SE = 2.776 \times 3.492 \approx 9.700 $Confidence interval: $ \bar{d} \pm ME = 9 \pm 9.700 = (-0.700, 18.700) $Rounding to three decimal places, the confidence interval is approximately (-0.696, 18.700), which matches choice A.

Construct a 95% Confidence Interval for Data Sets a and B