ANOVA table $\begin{array}{lrrrr} \hline\text { Source } & S S & d f & M S & F &p \text {-value }\ \hline \text { Treatment } & 6.000 & 3 & 1.9998 & 18.85&3.4E-05\ \hline\text { Error } & 1.486 & 14 & 0.1061 & \ \hline\text { Total } & 7.485 & 17 & & \end{array}$ Post hoc analysis Tukey simultaneous comparison t-values $ ( $ d.f. $ =14) $ $\begin{array}{ccccc} &\text { Brand 3 } \text { Brand 2 } \text { Brand 4 } \text { Brand 1 } \ &\quad1.40 \quad 2.28 \quad 2.58 \quad 2.95\ \begin{array}{ll} \text { Brand 3 } & 1.40 \ \text { Brand 2 } & 2.28 \ \text { Brand 4 } & 2.58 \ \text { Brand 1 } & 2.95 \end{array}&\begin{array}{|l|l|l|l|} \hline & & & \ \hline 4.27 & & & \ \hline 5.38 & 1.35 & & \ \hline 7.09 & 3.07 & 1.63 &\quad\quad \ \hline \end{array} \end{array}$ The Excel/Mega-Stat output given above summarizes the results of a one-way analysis of variance in an attempt to compare the performance characteristics of four brands of vacuum cleaners. The response variable is the amount of time it takes to clean a specific size room with a specific amount of dirt. -Use the information above and determine an individual 95% confidence interval confidence interval for $\mu _ { 1 } - \mu _ { 2 }$ )The mean and sample sizes for brand 1 and brand 2 are as follows: $\bar { x } _ { 1 }$ = 2)95, $\bar { x } _ { 2 }$ = 2)28,n₁ = 4 and n₂ = 5. A)2013 to 1.1387 B)2852 to 1.0547 C)-.228 to 1.5680 D)4515 to .8885 E)It cannot be determined with the information given.

Question

Quizplus · Accepted Answer

The correct answer is A because the confidence interval for $\mu _ { 1 } - \mu _ { 2 }$ is given by:$\bar { x } _ { 1 } - \bar { x } _ { 2 } \pm t _ { \alpha / 2 , \nu } \sqrt { \frac { s _ { 1 } ^ { 2 } } { n _ { 1 } } + \frac { s _ { 2 } ^ { 2 } } { n _ { 2 } } }$where $\alpha$ is the level of significance, $\nu$ is the degrees of freedom, $s_1^2$ and $s_2^2$ are the variances of the two samples, and $n_1$ and $n_2$ are the sample sizes.In this case, $\alpha = 0.05$, $\nu = 14$, $s_1^2 = 0.1061$, $s_2^2 = 0.1061$, $n_1 = 4$, and $n_2 = 5$. Plugging these values into the formula, we get:$\bar { x } _ { 1 } - \bar { x } _ { 2 } \pm t _ { 0.025 , 14 } \sqrt { \frac { 0.1061 } { 4 } + \frac { 0.1061 } { 5 } }$$= 2.95 - 2.28 \pm 2.145 \sqrt { 0.0265 + 0.0212 }$$= 0.67 \pm 0.247$$= (0.423, 0.917)$Therefore, the 95% confidence interval for $\mu _ { 1 } - \mu _ { 2 }$ is (0.423, 0.917).

ANOVA Table Post Hoc Analysis Tukey Simultaneous Comparison T-Values ( ( (

ANOVA Table Post Hoc Analysis
Tukey Simultaneous Comparison T-Values $($