Question

28. The following is sample information Test the hypothesis that the treatment means are equal. Treatment Treatment, Treatment 9 13 10 7 20 9 11 14 15 9 13 14 12 15 10 The null hypothesis is H.-H2 H3. What is your conclusion regarding He? ASIN B0040.18.5.700C QUESTION 29 29. to illustrate the effects of drivind under the influence of alcohol, police officer brought a Dur simulator to a high school. Student reaction time in an emergency was measured with unimpared vision and also while wearing special goggles to site the effects of alcohol on vision with the following result: Normal 4.47, 424,458,465,431,42 455.5.0.4.79 Impared: 5.77.5.67.551.5.32.583.5.49.5.23.5.61.5.63 is there evidence to suggest that there is a difference of 0.05 level of significance State Ho and Me Ho Ho Hi versus My 2 H2 OR MOH-H2 versus My Oc. Moritz versus MSW2 M 2 W versus Hawi su

Answer 1

Answer:

Given that,

$\rightarrow$ The following is a sample test of the hypothesis that means are equal.

Treatment 1	Treatment 2	Treatment 3
9	13	10
7	20	9
11	14	15
9	13	14
12		15
10

Claim:

Treatments means are equal.

The null and alternative hypothesis is:

$H_0:\mu1=\mu2=\mu3$

$H_1$: At least one of the mean is different.

Level of significance = 0.05

By using excel we can solve this question easily:

First, enter all data into excel.

Click on Data -------> Data Analysis--------> Anova: Single-Factor-----> Input Data -------> Select all data -------> Output Range : Select any empty cell --------> ok

We get,

7 2 Cla New Microsoft Office Excel Worksheet - Microsoft Excel | Home Insert Page Layout Formulas Data Review View Team Σ AutoSum Calibri - 11 AA Wrap Text General Fill - * Cut E Copy Paste Format Painter Clipboard BIU Merge & Center % -0.00 .00.0 Insert Delete Format Conditional Format Cell Formatting as Table - Styles Styles 2 Clear Sort & Find & Filter Select Editing Font Alignment Number Cells K40 V A B C D E F G н J K L N M - 0 P Q R S 21 22 23 Anova: Single Factor 24 25 26 Treatmen Treatmen Treatment3 9 13 10 7 20 9 11 14 15 9 13 14 12 15 10 Count 27 SUMMARY Groups Treatment1 Treatment2 Treatment3 6 Sum Average Variance 58 9.666667 3.066667 60 15 11.33333 63 12.6 8.3 28 4 29 5 30 31 32 33 df ANOVA Source of Variation Between Groups Within Groups SS 70.4 82.53333 F 5.118 MS 2 35.2 12 6.877778 P-value 0.0247 34 Fcrit 3.885 35 36 37 Total 152.9333 14 38 39 40 41 42 43 44 Activate Windows Go to PC settings to activate Windows. 45 KI Sheet1 Ready Sheet2 Sheet3 IIII OU 100% + e Σα O R M O 0 % 5:31 AM 01/17/2020

Test statistic = F = 5.118

Critical value = 3.885

The p-value= 0.0247

Since the p-value < 0.05

So, we reject H0.

Test statistic > Critical value we reject the null hypothesis.

Conclusion:

Not all means are equal.

Answer: Option A

Note:

** As per Chegg guidelines we should solve only the first question. So I have done it.
For the other question please post differently mentioning your requirement.