### Hypothesis Testing Example**Objective:**Test the hypothesis that the average content of containers of a particular lubricant is 10 liters.**Sample Data:**Content measurements from a random sample of 10 containers are:- 10.2, 9.7, 10.1, 10.3, 10.1, 9.8, 9.9, 10.4, 10.3, 9.8 liters**Parameters:**- **Null Hypothesis (H0):** The average content is 10 liters.- **Alternative Hypothesis (H1):** The average content is not 10 liters.- **Significance Level:** 0.01**Assumptions:**- The distribution of the contents is normal.**Methodology:**- Calculate the test statistic for the sample data.- Use the P-value to corroborate your decision on whether to reject or fail to reject the null hypothesis.

Name: ### Hypothesis Testing Example**Objective:**Test the hypothesis that
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Description: ### Hypothesis Testing Example**Objective:**Test the hypothesis that the average content of containers of a particular lubricant is 10 liters.**Sample Data:**Content measurements from a random sample of 10 containers are:- 10.2, 9.7, 10.1, 10.3, 10.

Given : A random sample of 10 containers of a particular lubricant is provided . The distribution of contents is normal . Calculate Mean & Standard deviation : The sample size is n = 10n=10. The provided sample data along with the data required to compute the sample mean X¯ and sample variance s2 are shown in the table below: X X2 10.2 104.04 9.7 94.09 10.1 102.01 10.3 106.09 10.1 102.01 9.8 96.04 9.9 98.01 10.4 108.16 10.3 106.09 9.8 96.04 Sum = 100.6 1012.58 The sample mean X¯ is computed as follows : Also, the sample variance s2 is , Therefore, the sample standard deviation s is , The calculated sample mean is x¯=10.06 and the sample standard deviation is s=0.246, and the sample size is n=10. Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho : Average content of containers of particular lubricant 10 liters . ie. μ = 1010 Ha : Average content of containers of particular lubricant is not equal to 10 liters . i.e. μ ≠ 1010 This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. Rejection Region Based on the information provided, the significance level is \alpha = 0.01α=0.01, and the critical value for a two-tailed test is tc=3.25. The rejection region for this two-tailed test is R = \{t: |t| > 3.25\}R={t:∣t∣>3.25} Test Statistics The t-statistic is computed as follows : Test statistic = 0.771 Decision about the null hypothesis Since it is observed that ∣t∣=0.771≤tc=3.25, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p=0.4603, and since p=0.4603 ≥ 0.01, it is concluded that the null hypothesis is not rejected. Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the average content of containers of particular lubricant is not equal to 10 liters , at the 0.01 significance level.

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Statistics

Test the hypothesis that the average content of containers of a particular lubricant is 10 liters if the contents of a random sample of 10 containers are {10.2, 9.7, 10.1, 10.3, 10.1, 9.8, 9.9, 10.4, 10.3, 9.8}. Use a 0.01 level of significance and assume that the distribution of contents is normal. Additionally, use P-value to corroborate your decision.

Test the hypothesis that the average content of containers of a particular lubricant is 10 liters if the contents of a random sample of 10 containers are {10.2, 9.7, 10.1, 10.3, 10.1, 9.8, 9.9, 10.4, 10.3, 9.8}. Use a 0.01 level of significance and assume that the distribution of contents is normal. Additionally, use P-value to corroborate your decision.

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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