Question

The following data represent the results from an independent-measures study comparing two treatment conditions. Treatment Treatment One Two 2.5 7.2 5.2 7.6 5.8 4.4 6.3 4.5 4.7 5.2 4.8 3 Using technology, run the One-way ANOVA Test for this data: F-ratio: p-value: Now, run the Two Independent Sample t test on the same data: Note: Do this with "pooled variances" since one assumption we make with ANOVA is that the variances for each group are equal. t-statistic: p-value: Do you notice anything interesting about the p-values? You might also note that the F Test Statistic is the square of of the t test statistic! F = ť?

Answer 1

1.

ANOVA-1 WAY:

The following table is obtained:

	Group 1	Group 2
	7.2	2.5
	5.2	5.8
	7.6	4.4
	6.3	4.5
	5.2	4.7
	4.8	3
Sum =	36.3	24.9
Average =	6.05	4.15
\sum_i X_{ij}^2 =∑i​Xij2​=	226.41	110.59
St. Dev. =	1.166	1.205
SS =	6.795	7.255
n =	6	6

The total sample size is N=12. Therefore, the total degrees of freedom are:

dftotal​=12−1=11

Also, the between-groups degrees of freedom are df_{between} = 2 - 1 = 1dfbetween​=2−1=1, and the within-groups degrees of freedom are:

dfwithin​=dftotal​−dfbetween​=11−1=10

First, we need to compute the total sum of values and the grand mean. The following is obtained

∑​Xij​=36.3+24.9=61.2

Also, the sum of squared values is

∑​Xij2​=226.41+110.59=337

Based on the above calculations, the total sum of squares is computed as follows

$SS_{total} = \sum_{i,j} X_{ij}^2 - \frac{1}{N} \left(\sum_{i,j} X_{ij}\right)^2 = 337 - \frac{ 61.2^2}{ 12} = 24.88$

The within sum of squares is computed as shown in the calculation below:

$SS_{within} = \sum SS_{within groups} = 6.795+7.255 = 14.05$

$MS_{between} = \frac{SS_{between}}{df_{between}} = \frac{ 10.83}{ 1} = 10.83$

$MS_{within} = \frac{SS_{within}}{df_{within}} = \frac{ 14.05}{ 10} = 1.405$

$F = \frac{MS_{between}}{MS_{within}} = \frac{ 10.83}{ 1.405} = 7.70$

F=7.70

p-value=0.0196

2. t-test

First we calculate mean and standard deviation of both the samples:

	Treatment 1	Treatment 12
	7.2	51.84
	5.2	27.04
	7.6	57.76
	6.3	39.69
	5.2	27.04
	4.8	23.04
Sum =	36.3	226.41

The sample mean \bar XXˉ is computed as follows:

$\bar X = \frac{1}{n} \sum_{i=1}^n X_i = \frac{ 36.3}{ 6} = 6.05$

Also, the sample variance s2 is

$s^2 = \frac{ 1}{n-1}\left(\sum_{i=1}^n X_i^2 - \frac{1}{n}\left(\sum_{i=1}^n X_i\right)^2 \right) = \frac{ 1}{ 6-1}\left( 226.41 - \frac{ 36.3^2}{ 6} \right) = 1.359$

Therefore, the sample sandartd deviation s is

$s = \sqrt{s^2} = \sqrt{ 1.359} = 1.166$

	treatment 2	treatment 22
	2.5	6.25
	5.8	33.64
	4.4	19.36
	4.5	20.25
	4.7	22.09
	3	9
Sum =	24.9	110.59

The sample mean Xˉ is computed as follows:

$\bar X = \frac{1}{n} \sum_{i=1}^n X_i = \frac{ 24.9}{ 6} = 4.15$

Also, the sample variance s2 is

1 1 24.92 52 (Σ' ! (Σ») ) -:-( (1 110.59 - 1.451 η – 6

Therefore, the sample sandartd deviation s is

$s = \sqrt{s^2} = \sqrt{ 1.451} = 1.205$

HENCE:

Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

t X1 - X2 (ni-1)s +(n2-1)33 ni+n2-2 + 12

t= 2.776 6.05 – 4.15 (6-1)1.1662+(6-1)1.2052 6+6-2 (+)

t=2.776

p-value=0.0196

1.We observe that the p-values remains same in both the cases p-value=0.0196

2.t=2.776

$t^2=2.776^2=7.70$ =F

Hence shown.

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