Hypothesis testing part iii for difference of means
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This document discusses hypothesis testing for the difference between means. It provides three examples that demonstrate how to perform hypothesis tests to compare the means of two samples. The examples show how to: 1) State the null and alternative hypotheses, 2) Determine the test statistic and critical region based on the hypotheses, level of significance and test assumptions, 3) Perform calculations to obtain the test statistic value, and 4) Make a conclusion about whether to reject or fail to reject the null hypothesis based on comparing the test statistic to the critical region.
Hypothesis testing part iii for difference of means
- 1. HYPOTHESIS TESTING PART-III DIFFERENCE OF MEANS NADEEM UDDIN ASSOCIATE PROFESSOR OF STATISTICS
- 2. Hypothesis Test for the difference between means. Statisticians follow a formal process to determine whether to reject a null hypothesis, based on sample data. This process called hypothesis testing. 1. State the hypotheses. This involves stating the null and alternative hypotheses. The hypotheses are stated in such a way that they are mutually exclusive. That is, if one is true, the other must be false. Set Null hypothesis Alternative hypothesis Number of tails 1 μ1 - μ2 = d μ1 - μ2 ≠ d 2 2 μ1 - μ2 > d μ1 - μ2 < d 1 3 μ1 - μ2 < d μ1 - μ2 > d 1
- 3. 2. Level of significance: α = 0.01, 0.05 or any given value 2 21 2 1 2 2 2 1 2 1 2 1 2 1. Z= and known X X when n n 2 21 2 1 2 2 2 1 2 1 2 1 2 1, 22. Z= and unknown and n 30 X X when n n n S S 2 21 2 1 2 1 2 1 2 1, 23. t = and unknown and n < 1 1 30 p X X when n n n S 3. Test Statistic
- 4. 4. Critical Region: The set of values outside the region of acceptance is called the region of rejection. If the test statistic falls within the region of rejection, the null hypothesis is rejected. In such cases, we say that the hypothesis has been rejected at the α level of significance. The following steps are use to find the critical region. For Test statistic (1) and (2) Z > Zα/2 and Z< - Zα/2 When H1: μ1 - μ2 ≠ d Z > Zα When H1: μ1 - μ2 > d Z< - Zα When H1: μ1 - μ2 < d For Test statistic (3) t > tα/2,υ and t < - tα/2,υ When H1: μ1 - μ2 ≠ d t > tα, υ When H1: μ1 - μ2 > d t < - tα, υ When H1: μ1 - μ2 < d ;Where v = n1+n2 - 2
- 5. 5. Computation: Find the value of the test statistic 6. Conclusion: If the calculated value of test statistic falls in the area of rejection, we reject the null hypothesis otherwise accept it.
- 6. Test Concerning Double Means Example-1: Two independent samples of observations were collected for the first sample of 60 elements, the mean was 86 and the standard deviation 6. The second sample of 75 elements had a mean of 82 and a standard deviation of 9. Using α=0.01, test whether the two samples can reasonably be considered to have come from populations with the same mean.
- 7. Solution: n1=60 n2=75 s1=6 s2=9 𝑥1=86 𝑥2=82 α = 0.01 1. Hypothesis H0: 𝜇1 − 𝜇2 = 0 H1: 𝜇1 − 𝜇2 ≠ 0 2. Level of significance α = 0.01 3. Test statistic 1 2 1 2 2 2 1 2 1 2 X X z S S n n
- 8. 4. Critical Region In case of two tail test i.e. H1 𝑖𝑠 ≠. Reject H0, if 𝑍 𝑐𝑎𝑙 ≤ −𝑍𝑡𝑎𝑏 or 𝑍 𝑐𝑎𝑙 ≥ 𝑍𝑡𝑎𝑏. Where 𝑍𝑡𝑎𝑏 = 𝑍 𝛼 2 = 𝑍0.01 2 = 𝑍0.005 = 2.58 𝑍 𝑐𝑎𝑙 ≤ −2.58 or 𝑍 𝑐𝑎𝑙 ≥ 2.58. (Using inverse area of normal table) 5. Computation 2 2 (86 82) (0) 6 9 60 75 Z =Zcal= 3.09 6. Conclusion: Reject H0. 2.58 –0– 2.58
- 9. Example-2: A manufacturer claims that the average tensile strength of thread A exceeds the average tensile strength of thread B by at least 12kg. To test this claim 50 pieces of each thread are tested under similar condition. Type A thread had an average tensile strength of 80kg with a standard deviation of 5kg. While type B thread had an average tensile strength of 70kg. With a standard deviation of 4kg. Test the manufacturer’s claim using 0.01 level of significance.
- 10. Solution: n1 = 50 n2 = 50 s1 = 5 s2 = 4 𝑥1 = 80 𝑥2 = 70 α = 0.01 1. Hypothesis H0: 𝜇1 − 𝜇2 ≥ 12 H1: 𝜇1 − 𝜇2 < 12 2. Level of significance α = 0.01 3. Test statistic 1 2 1 2 2 2 1 2 1 2 X X z S S n n
- 11. 4. Critical Region 5. Computation 2 2 (80 70) (12) 2.21 5 4 50 50 Z 6. Conclusion: Accept H0. 2.33 –0– In case of lower tail test i.e. H1 𝑖𝑠 ˂. Reject H0, if 𝑍 𝑐𝑎𝑙 ≤ −𝑍𝑡𝑎𝑏 Where 𝑍𝑡𝑎𝑏 = 𝑍 𝛼 = 𝑍0.01 = −2.33 𝑍 𝑐𝑎𝑙 ≤ −2.33 (Using inverse area of normal table)
- 12. Example-3: A course in mathematics is taught to 12 students by the conventional classroom procedure. A second group of 10 students was given the same course by means of programmed materials. At the end of the semester the same examination was given each group. The 12 students meeting in the classroom made an average grade of 85 with a standard deviation of 4, while the 10 students using programmed materials made an average of 81 with a standard deviation of 5. Test the hypothesis that the two methods of learning are equal using a 0.10 level of significance. Assume the populations to be approximately normal with equal variances.
- 13. Solution: n1 = 12 n2 = 10 s1 = 4 s2 = 5 𝑥1 = 85 𝑥2 = 81 α = 0.10 1 .Hypothesis H0: 𝜇1 − 𝜇2 = 0 H1: 𝜇1 − 𝜇2 ≠ 0 2. Level of significance α = 0.10 3. Test statistic 1 2 1 2 1 2 1 1 p X X t n n S
- 14. 1.725 –0– 1.725 4. Critical Region In case of two tail test i.e. H1 𝑖𝑠 ≠. Reject H0, if 𝑡 𝑐𝑎𝑙 ≤ −𝑡𝑡𝑎𝑏 or 𝑡 𝑐𝑎𝑙 ≥ 𝑡𝑡𝑎𝑏. Where 𝑡𝑡𝑎𝑏 = 𝑡 𝛼 2 ,(𝑛1+𝑛2−2) = 𝑡0.10 2 ,(12+10−2) = 𝑡0.05,20 = 1.725 𝑡 𝑐𝑎𝑙 ≤ −1.725 or 𝑡 𝑐𝑎𝑙 ≥ 1.725.
- 15. 5. Computation 2 2 1 1 2 2 1 2 1 1 2 n s n s sp n n 2 2 12 1 10 1 12 10 2 (4) (5) 4.478 85 81 0 1 1 4.478 12 10 2.07cal sp t t 6. Conclusion: Reject H0.