**Question #6 for Students with Last Names Beginning with R-Z:**The following scores are from an independent-measures study comparing two treatment conditions:1. Use an independent-measures t-test with α = .05 to determine whether there is a significant mean difference between the two treatments.2. Use an ANOVA with α = .05 to determine whether there is a significant mean difference between the two treatments. You should find that F = t².**Data Table:**| Treatment I | Treatment II ||-------------|--------------|| 10 | 7 || 8 | 4 || 7 | 9 || 9 | 3 || 13 | 7 || 7 | 6 || 6 | 10 || 12 | 2 |**Summary Statistics:**- Total number of scores ($N$) = 16- Grand total of scores ($G$) = 120- Sum of squares of scores ($\Sigma X^2$) = 1036In this exercise, you will perform both a t-test and an ANOVA to analyze the data from two different treatment groups, determining if there is a statistically significant difference between them.

From Given Data: x¯1=∑xn=10+8+7+9+13+7+6+128=9x¯2=∑xn=7+4+9+3+7+6+10+28=6…

Answered: **Question #6 for Students with Last…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Expert Solution

Step 1

From Given Data:

${\bar{x}}_{1} = \frac{\sum_{} x}{n} = \frac{10 + 8 + 7 + 9 + 13 + 7 + 6 + 12}{8} = 9 {\bar{x}}_{2} = \frac{\sum_{} x}{n} = \frac{7 + 4 + 9 + 3 + 7 + 6 + 10 + 2}{8} = 6$

$s_{1}^{2} = \frac{\sum_{}^{} {(x - \bar{x})}^{2}}{n - 1} = \frac{44}{7} = 6.2857 s_{2}^{2} = \frac{\sum_{}^{} {(x - \bar{x})}^{2}}{n - 1} = \frac{56}{7} = 8$

Calculating the pooled variance

$s_{p}^{2} = \frac{s_{1}^{2} (n_{1} - 1) + s_{2}^{2} (n_{2} - 1)}{n_{1} + n_{2} - 2} = \frac{6.2857 \times 7 + 8 \times 7}{8 + 8 - 2} = 7.1429$

Step 2

1) Independent Measure t-test

H0: $μ_{1} = μ_{2}$

HA: $μ_{1} \neq μ_{2}$

Test Statistics:

$\frac{{\bar{x}}_{1} - {\bar{x}}_{2}}{s_{p} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}} = \frac{9 - 6}{\sqrt{7.1429} \times \sqrt{\frac{1}{8} + \frac{1}{8}}} = 2.245$

Critical Value for two tailed t-test:

Degree of freedom=0.05

Using t-distribution table

Critical t-value=2.145

As test statistic value (2.245) is greater than the critical value (2.145),

we have sufficient evidence at 5% significance level to reject null hypothesis

Therefore we can conclude that there is a significant mean difference between the two treatments.

Step 3

2) ANOVA test

$S o u r c e o f v a r i a t i o n t o t a l (S S_{T}) = \sum_{} X^{2} - \frac{{(\sum_{}^{} x_{1} + \sum_{}^{} x_{2})}^{2}}{n_{1} + n_{2}} = 1036 - \frac{{(72 + 48)}^{2}}{8 + 8} = 136$

$S o u r c e o f v a r i a t i o n b e t w e e n t r e a t m e n t s (S S_{B}) = (\frac{{(\sum_{}^{} x_{1})}^{2}}{n_{1}} + \frac{{(\sum_{}^{} x_{2})}^{2}}{n_{2}}) - \frac{{(\sum_{}^{} x_{1} + \sum_{}^{} x_{2})}^{2}}{n_{1} + n_{2}} = (\frac{72^{2}}{8} + \frac{48^{2}}{8}) - \frac{{(72 + 48)}^{2}}{8 + 8} = 36$