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. $$\begin{array} { | l | l l l l l | }
\hline \mathrm { A } & 5.0 & 5.1 & 4.6 & 3.5 & 6.0 \
\hline \mathrm { B } & 4.7 & 3.9 & 4.2 & 4.2 & 3.7 \
\hline
\end{array}$$
A) $- 0.31

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. $$\begin{array} { | l | l l l l l | } 
\hline \mathrm { A } & 5.0 & 5.1 & 4.6 & 3.5 & 6.0 \
\hline \mathrm { B } & 4.7 & 3.9 & 4.2 & 4.2 & 3.7 \
\hline
\end{array}$$ 
A) $- 0.31 < \mu _ { d } < 1.71$
B) $- 0.07 < \mu _ { d } < 1.47$
C) $0.22 < \mu _ { d } < 7.48$
D) $- 0.37 < \mu _ { d } < 1.77$

Quizplus · Accepted Answer

The formula to calculate the confidence interval for population mean of paired data is given by:

$$\bar{d} \pm t_{\frac{\alpha}{2}, n-1}\frac{s_d}{\sqrt{n}}$$

where $\bar{d}$ is the sample mean of paired differences, $s_d$ is the sample standard deviation of paired differences, $n$ is the sample size and $t_{\frac{\alpha}{2}, n-1}$ is the t-value for a two-tailed test with degrees of freedom $(n-1)$ at the confidence level $(1-\alpha)$.

Here, we have n=5 paired data, so $(n-1)=4$ degrees of freedom.

First, we need to calculate $\bar{d}$ and $s_d$ as follows:

$$\bar{d} = \frac{\sum_{i=1}^{n} d_i}{n} = \frac{(5.0-4.7)+(5.1-3.9)+(4.6-4.2)+(3.5-4.2)+(6.0-3.7)}{5} = 0.9$$

$$s_d = \sqrt{\frac{\sum_{i=1}^{n} (d_i-\bar{d})^2}{n-1}} = \sqrt{\frac{(5.0-0.9)^2+(5.1-0.9)^2+(4.6-0.9)^2+(3.5-0.9)^2+(6.0-0.9)^2}{4}} = 1.55$$

Using a t-distribution table or calculator with 4 degrees of freedom and a 90% confidence level, we find that the t-value is 2.132.

Substituting the values into the formula, we get:

$$0.9 \pm 2.132\frac{1.55}{\sqrt{5}}$$

Simplifying the expression, we get:

$$- 0.37 < \mu _ { d } < 1.77$$

Therefore, the correct choice is D, i.e. $- 0.37 < \mu _ { d } < 1.77$.

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

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