Chapter 15, Problem 1PP

a.

To determine

To provide: The null and alternate hypothesis for ANOVA.

Answer to Problem 1PP

The hypotheses are:

Null Hypothesis: There is no significant difference in the difference between the amount of nectar consumed from caffeine feeders and that removed from control feeders at different concentrations of caffeine.

Alternate Hypothesis: There is at least one concentration group which is significantly different from in the difference between the amount of nectar consumed from caffeine feeders and that removed from control feeders at different concentrations of caffeine.

Explanation of Solution

Given:

The data of four different concentrations of caffeine with difference between the amount of nectar consumed from caffeine feeders and that are removed from control feeders as shown below

50 ppm caffeine	100 ppm caffeine	150 ppm caffeine	200 ppm caffeine
-0.4	0.01	0.65	0.24
0.34	-0.39	0.53	0.44
0.19	-0.08	0.39	0.13
0.05	-0.09	-0.15	1.03
-0.14	-0.31	0.46	0.05

Calculation:

The objective of the ANOVA is to check if there is any significant difference in the difference between the amount of nectar consumed from caffeine feeders and that removed from control feeders at different concentrations of caffeine.

Null Hypothesis: There is no significant difference in the difference between the amount of nectar consumed from caffeine feeders and that removed from control feeders at different concentrations of caffeine.

Alternate Hypothesis: There is at least one concentration group which is significantly different from in the difference between the amount of nectar consumed from caffeine feeders and that removed from control feeders at different concentrations of caffeine.

b.

To determine

To Calculate: The summary statistics for each group.

Answer to Problem 1PP

The summary statistics of each group are shown below

  $\boxed{\begin{array}{ccccc}Groups& count\left(n_i\right)& Mean\left({\stackrel{\_\_}{Y}}_i\right)& Variance& \text{Standard deviation}\left(s_i\right)\\ \begin{array}{l}50\text{ ppm Caffeine}\\ \text{100 ppm Caffeine}\end{array}& \begin{array}{l}5\\ 5\end{array}& \begin{array}{l}0.008\\ -0.172\end{array}& \begin{array}{l}0.08337\\ 0.02872\end{array}& \begin{array}{l}0.28874\\ 0.16947\end{array}\\ 150\text{ ppm Caffeine}& 5& 0.376& 0.09568& 0.30932\\ 200\text{ ppm Caffeine}& 5& 0.378& 0.15427& 0.39278\end{array}}$

Explanation of Solution

Calculation:

In excel enter the data of four different concentrations of caffeine with difference between the amount of nectar consumed from caffeine feeders and that are removed from control feeders. Click on Data → Data Analysis → ANOVA Single factor and enter the data we get the following summary statistics for each concentration.

  

The standard deviation for given variance is calculated as shown below

  $\begin{array}{c}\text{For 50 ppm caffeine}\\ \text{Standard deviation }=\sqrt{Variance}\\ =\sqrt{0.08337}\\ =0.28874\\ \text{For 100 ppm caffeine}\\ \text{Standard deviation }=\sqrt{Variance}\\ =\sqrt{0.02872}\\ =0.16947\\ \text{For 150 ppm caffeine}\\ \text{Standard deviation }=\sqrt{Variance}\\ =\sqrt{0.09568}\\ =0.30932\\ \text{For 200 ppm caffeine}\\ \text{Standard deviation }=\sqrt{Variance}\\ =\sqrt{0.15427}\\ =0.39278\end{array}$

c.

To determine

To calculate: The ANOVA table

Explanation of Solution

Calculation:

In excel enter the data of four different concentrations of caffeine with difference between the amount of nectar consumed from caffeine feeders and that are removed from control feeders. Click on Data → Data Analysis → ANOVA Single factor and enter the data.

The obtained output is

  

d.

To determine

To calculate: The Mean Square error in ANOVA.

Answer to Problem 1PP

The Mean square error $M{S}_{error}=0.09051$

Explanation of Solution

Calculation:

With reference to ANOVA table the mean square error is the Mean square within groups

  $M{S}_{error}=0.09051$

e.

To determine

To calculate: The degree of freedom with respect to error or within groups.

Answer to Problem 1PP

The degree of freedom with respect to within groups is 16.

Explanation of Solution

Calculation:

The degree of freedom for within groups with respect to the ANOVA table

  $Degreeoffreedomwithingroups=16$

f.

To determine

To Calculate: The grand mean of given data.

Answer to Problem 1PP

The grand mean of given data is 0.1475.

Explanation of Solution

The summary statistics of given data from ANOVA output is shown below

  

Calculation:

With respect to summary statistics of given data the grand mean is calculated as shown below

  $\begin{array}{c}\stackrel{\_\_}{Y}=\frac{{\displaystyle \sum _{i=1}^4{Y}_i}}{n}\\ \stackrel{\_\_}{Y}=\frac{0.04+(-0.172)+0.376+0.378}{5+5+5+5}\\ =\frac{2.95}{20}\\ =0.1475\end{array}$

g.

To determine

To Calculate: The group sum of squares

Answer to Problem 1PP

The group sum of squares is 1.331415

Explanation of Solution

Calculation:

With respect to the ANOVA table the sum of squares for between the groups is

  $Sumofsquaresbetweengroups=1.134415$

h.

To determine

To calculate: The group degree of freedom and group mean square.

Answer to Problem 1PP

The degree of freedom for group is 3 and the group mean square is 0.378138.

Explanation of Solution

Calculation:

With respect to the ANOVA table, the group degree of freedom is 3.

The mean square for group is $M{S}_{group}=0.378138$

i.

To determine

To calculate: The test statistic F.

Answer to Problem 1PP

The test statistic F is 3.238872

Explanation of Solution

Calculation:

With respect to ANOVA table the test statistic F is 3.238872.

j.

To determine

To calculate: The p value with respect to test statistic F and degree of freedom.

Answer to Problem 1PP

The p value with respect to test statistic F and degree of freedom is 0.023078.

Explanation of Solution

Calculation:

The p value with respect to test statistic F 3.238872 and degree of freedom 3, 16 is 0.023078.