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Formula: df = n 1 (for one-sample B. df = n -1 mean test) df = 16-1 df = (₁ + ₂) - 2 (for two- sample mean test) df = 15 C. The critical or tabular value (teritical). (Can be found on C. 2.947 the given table of Tabular Value of T)

Formula: df = n 1 (for one-sample B. df = n -1 mean test) df = 16-1 df = (₁ + ₂) - 2 (for two- sample mean test) df = 15 C. The critical or tabular value (teritical). (Can be found on C. 2.947 the given table of Tabular Value of T)

MATLAB: An Introduction with Applications
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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Formula: df = n 1 (for one-sample B. df = n-1 mean test) df = 16-1 df = (₁ + n₂) - 2 (for two- sample mean test) df = 15 C. The critical or tabular value (teritical). (Can be found on C. 2.947 the given table of Tabular Value of T) 4. Compute for the value of the Formula: test statistic (t computed) using the data from the population and the sample/s. Given: X-48.1 μ- n-29 s-0.9 Substitute the given to the formula and solve: 5. Make a decision. Decision: (Reject Ho and Accept Ha or Accept Ho and Reject Ha) 6. State the conclusion Conclusion: (x-µ)√n
Transcribed Image Text:Formula: df = n 1 (for one-sample B. df = n-1 mean test) df = 16-1 df = (₁ + n₂) - 2 (for two- sample mean test) df = 15 C. The critical or tabular value (teritical). (Can be found on C. 2.947 the given table of Tabular Value of T) 4. Compute for the value of the Formula: test statistic (t computed) using the data from the population and the sample/s. Given: X-48.1 μ- n-29 s-0.9 Substitute the given to the formula and solve: 5. Make a decision. Decision: (Reject Ho and Accept Ha or Accept Ho and Reject Ha) 6. State the conclusion Conclusion: (x-µ)√n
Given: sample size, n=29 sample mean, x=48.1 population standard deviation,a=0.9 population mean,µμ= 50 Significance level, a=0.01 1) Hypothesis test: The null and alternative hypothesis is Ho: The average weight of bag of chemical is equal to 50 kilograms. In symbol Ho: u = 50 Ha: The average weight of bag of chemical is not equal to 50 kilograms. In symbol Ha: u = 50 2) a) Significance level, a=0.01 The hypothesis test is two tailed test. Step 4 3)a) t test.: One sample mean. B) Degree of freedom df-n-1=29-1=28 C) critical value: critical value at a =0.01 and degree of freedom, df-28 is ta/2= +2.763 from t distribution table
Transcribed Image Text:Given: sample size, n=29 sample mean, x=48.1 population standard deviation,a=0.9 population mean,µμ= 50 Significance level, a=0.01 1) Hypothesis test: The null and alternative hypothesis is Ho: The average weight of bag of chemical is equal to 50 kilograms. In symbol Ho: u = 50 Ha: The average weight of bag of chemical is not equal to 50 kilograms. In symbol Ha: u = 50 2) a) Significance level, a=0.01 The hypothesis test is two tailed test. Step 4 3)a) t test.: One sample mean. B) Degree of freedom df-n-1=29-1=28 C) critical value: critical value at a =0.01 and degree of freedom, df-28 is ta/2= +2.763 from t distribution table
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