could you please explain step 3 how did they get 0.00001 Step 1From the provided information,Sample size (n) = 30%3DSample mean (X) = 111Sample standard deviation (s) = 1.7Since, the population standard deviation is unknown, therefore, the t distribution will be useStep 2The hypotheses are as follow:Ho:H = 120Ha:H # 120Confidence level = 95%Level of significance (a) = 1- 0.95 = 0.05The value of the test statistic can be obtained as:x µS |t =S111–1201.730= -28.9971The test statistic value is -28.9971.Step 3The degree of freedom = n - 1= 30 – 1= 29The p value of the test statistic at 29 degrees of freedom from the t value table is 0.00001 whichis less than level of significance therefore, the null hypothesis would reject and it can beconcluded that there is sufficient evidence to support the claim that the machine is notworking properly.

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Description: could you please explain step 3 how did they get 0.00001 Step 1From the provided information,Sample size (n) = 30%3DSample mean (X) = 111Sample standard deviation (s) = 1.7Since, the population standard deviation is unknown, therefore, the t distribu

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Step 1 From the provided information, Sample size (n) = 30 Sample mean (x) = 111 Sample standard deviation (s) = 1.7 Since, the population standard deviation is unknown, therefore, the t distribution will be use Step 2 The hypotheses are as follow: Ho:H = 120 HaiH # 120 Confidence level = 95% Level of significance (a) = 1- 0.95 = 0.05 The value of the test statistic can be obtained as: x - µ t = S

Step 1 From the provided information, Sample size (n) = 30 Sample mean (x) = 111 Sample standard deviation (s) = 1.7 Since, the population standard deviation is unknown, therefore, the t distribution will be use Step 2 The hypotheses are as follow: Ho:H = 120 HaiH # 120 Confidence level = 95% Level of significance (a) = 1- 0.95 = 0.05 The value of the test statistic can be obtained as: x - µ t = S

MATLAB: An Introduction with Applications

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ISBN:9781119256830

Author:Amos Gilat

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Topic Video

Question

could you please explain step 3 how did they get 0.00001

Step 1
From the provided information,
Sample size (n) = 30
%3D
Sample mean (X) = 111
Sample standard deviation (s) = 1.7
Since, the population standard deviation is unknown, therefore, the t distribution will be use
Step 2
The hypotheses are as follow:
Ho:H = 120
Ha:H # 120
Confidence level = 95%
Level of significance (a) = 1- 0.95 = 0.05
The value of the test statistic can be obtained as:
x µ
S

|
t =
S
111–120
1.7
30
= -28.9971
The test statistic value is -28.9971.
Step 3
The degree of freedom = n - 1= 30 – 1= 29
The p value of the test statistic at 29 degrees of freedom from the t value table is 0.00001 which
is less than level of significance therefore, the null hypothesis would reject and it can be
concluded that there is sufficient evidence to support the claim that the machine is not
working properly.

Expert Solution

Step 1

Given:

Test Statistics = -28.9971

degree of freedom =29

Obtained p value = 0.00001

Step by step

Solved in 2 steps

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