Looking for a detailed answer to the sub parts g and f please. Thanks. graphStyles6. Check out the frequency distributions for two treatment groups:Treatment 1ITreatment 2n = 9M= 8n = 9M= 13S = 1.22S = 1.220 1 2 3 4 5 67 89 10 11 12 13 14 15 16 17 18 19 20 21a. Write a table with observations for these samples.b. Using the table, explain (calculate) where the S=1.22 comes fromC. Conduct a two-tailed comparison of the two groups with a=.05. Pleaseremember to state your hypothesis.Now look at:Treatment 1ITreatment 2 On = 9M= 8S = 6.65n = 9M= 13S = 6.65235 67 89 10 11 12 13 14 15 l6 17 18 19 20 21d. Write a table with observations for these samples.e. Using the table, explain (calculate) where the S=6.65 comes fromConduct a two-tailed comparison of the two groups with a=.05.f.g. Now combine the observation from the tables mentioned in parts (a) and (d),so that are 18 observations for each Treatment. What is the s for treatment 1and treatment 2? What is M?h. Conduct a two-tailed comparison of the two groups with a-.05.

Answered: f. Conduct a two-tailed comparison of…

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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Looking for a detailed answer to the sub parts g and f please. Thanks.

graph
Styles
6. Check out the frequency distributions for two treatment groups:
Treatment 1I
Treatment 2
n = 9
M= 8
n = 9
M= 13
S = 1.22
S = 1.22
0 1 2 3 4 5 6
7 8
9 10 11 12 13 14 15 16 17 18 19 20 21
a. Write a table with observations for these samples.
b. Using the table, explain (calculate) where the S=1.22 comes from
C. Conduct a two-tailed comparison of the two groups with a=.05. Please
remember to state your hypothesis.
Now look at:
Treatment 1I
Treatment 2 O
n = 9
M= 8
S = 6.65
n = 9
M= 13
S = 6.65
2
3
5 6
7 8
9 10 11 12 13 14 15 l6 17 18 19 20 21
d. Write a table with observations for these samples.
e. Using the table, explain (calculate) where the S=6.65 comes from
Conduct a two-tailed comparison of the two groups with a=.05.
f.
g. Now combine the observation from the tables mentioned in parts (a) and (d),
so that are 18 observations for each Treatment. What is the s for treatment 1
and treatment 2? What is M?
h. Conduct a two-tailed comparison of the two groups with a-.05.

Expert Solution

Step 1

f. For the two-tailed comparison for two groups, the null and alternative hypothesis are given by :

$H_{o} : μ_{1} = μ_{2} H_{1} : μ_{1} \neq μ_{2}$

The test statistics is given by Students' t-distribution because population standard deviation is unknown.

$t = \frac{x_{1} - x_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}} ~ t_{n_{1} + n_{2} - 2}$

$\begin{array}{rcl} t & = & \frac{8 - 13}{\sqrt{\frac{6 . 65^{2}}{9} + \frac{6 . 65^{2}}{9}}} \\ = & \frac{- 5}{3.135} \\ = & - 1.595 \end{array}$

The two-tailed critical value for t-distribution at $α = 0.05$ with $(18 - 2) = 16$ degrees of freedom is $\pm 2.120$

The null hypothesis is rejected according to the decision rule when calculated test statistics is greater than the tabulated value and is accepted otherwise.

Since $- 1.595 < - 2.120$ , null hypothesis is accepted concluding that the average for the two groups are same.

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