Construct a confidence interval for µd, the mean of the differences d for the population of paired data. Assume that the population of paired differences is normally distributed.
-If $\mathrm { d } = 3.125 , \mathrm {~S} _ { \mathrm { d } } = 2.911$, and $\mathrm { n } = 8$, determine a 99 percent confidence interval for $\mu _ { \mathrm { d } }$.
A) $- 0.476

Question

Construct a confidence interval for µd, the mean of the differences d for the population of paired data. Assume that the population of paired differences is normally distributed.
-If $\mathrm { d } = 3.125 , \mathrm {~S} _ { \mathrm { d } } = 2.911$, and $\mathrm { n } = 8$, determine a 99 percent confidence interval for $\mu _ { \mathrm { d } }$.
A) $- 0.476 < \mu _ { \mathrm { d } } < 6.726$
B) $1.851 < \mu _ { \mathrm { d } } < 6.726$
C) $1.851 < \mu _ { \mathrm { d } } < 4.399$
D) $- 0.476 < \mu _ { \mathrm { d } } < 4.399$

Quizplus · Accepted Answer

To construct a 99% confidence interval for the mean difference $\mu_d$, we use the formula: $\bar{d} \pm t_{\alpha/2} \times \frac{S_d}{\sqrt{n}}$, where $\bar{d} = 3.125$, $S_d = 2.911$, and $n = 8$. The degrees of freedom (df) is $n - 1 = 7$. For a 99% confidence interval and 7 degrees of freedom, the $t_{\alpha/2}$ value is approximately 3.499 (from t-distribution tables). Plugging in the values, we calculate the margin of error and the confidence interval, which results in $-0.476 < \mu_d < 6.726$.

Construct a Confidence Interval for µD, the Mean of the Differences