Statistics Presentation week 7

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Statistics Presentation week 7

  1. 1. STATISTICS & PROBABILITYHypothesis Testing
  2. 2. What is a Hypothesis?I assume the mean • an assumption about GPA of this class the population is 3.5! parameter • an educated guess about the population parameter
  3. 3. Hypotheses Testing: This is the process of making an inference or generalization onpopulation parameters based on the results of the study on samples. Reject? Accept? Statistical Hypotheses: It is a guess or prediction made by the researcher regarding the possible outcome of the study.
  4. 4. Hypotheses Testingis deciding between what is REALITY and what is COINCIDENCE!
  5. 5. Types of Statistical Hypotheses Null Hypothesis (Ho): is always hoped to be rejected Always contains “=“ signAlternative Hypothesis (Ha):•Challenges Ho•Never contains “=“ sign•Uses “< or > or ≠ “•It generally represents the idea which the researcher wants to prove.
  6. 6. The Null Hypothesis: HoEx. Ho: The average GPA of this class is 3.5 µ = 3.5 H0: The Alternative Hypothesis: Ha Ha: The average GPA of this class is a) higher than 3.5 (Ha: µ > 3.5) b) lower than 3.5 (Ha: µ < 3.5) c) not equal to 3.5 (Ha:µ ≠ 3.5)
  7. 7. Types of Hypotheses Tests1. One-tailed left directional test – this is used if Ha uses < symbol Critical value isα = 0.05 obtained Acceptance from the table region Area = 0.05 Rejection region
  8. 8. Types of Hypotheses Tests 2. One-tailed right directional test – this is used if Ha uses > symbol Critical value isα = 0.05 Acceptance obtained from the table region Area= 0.05 Rejection region
  9. 9. Types of Hypotheses Tests3. Two-tailed test: Non-directional – this is used if Ha uses ≠ symbol Critical value is α = 0.05/2 obtained Acceptance from the table region Area=.025 Area=.025 Rejection region Rejection region
  10. 10. Level of Significance, α and the Rejection Regionα 0.05, = means the probability of being right is 95% , and the probability of being wrong is 5%. So what is α 0.01? = Rejection region Acceptance Area is 0.05 Region α = 0.05 .
  11. 11. Level of Significance, α and the Rejection Regionα= 0.01, means the researcher is taking a 1% risk of being wrong and a 99% risk of being right. So, what is α = 0.05? Rejection region Area is 0.01 Acceptance Region α = 0.01
  12. 12. Level of Significance, α and the Rejection Regionα = 0.05 means the probability of committing Type I error is 5%. α = 0.05, since it is 2-T, then α = 0.05/2= 0.025 Acceptance region Area=.025 Area=.025 Rejection region Rejection region
  13. 13. Level of Significance, α and the Rejection RegionTo summarize:α= 0.05, means the probability of being right is 95% and the probability of being wrong is 5%. So what is α 0.01? =α= 0.01, means the researcher is taking a 1% risk of being wrong and a 99% risk of being right. So, what is α = 0.05?α = 0.05 means the probability of committing Type I error is5%. α So what is = 0.01?
  14. 14. Errors in Hypothesis Testing Ho: ERAP is not guiltyType I (α error ) ErrorsRejecting a true If the court convicts Ho! ERAP, when in fact he is not guilty, the Errors in Decisions court commits Type I error! Errors in Conclusions Type I is the same as the α or the level of significance. © 1984-1994 T/Maker Co.
  15. 15. Errors in Hypothesis TestingType II (β error ) ErrorsAccepting a false Ho: ERAP is not guilty Ho! If the court acquits Errors in Decisions ERAP, when in fact he is guilty, the court commits Type II error! Errors in Conclusions
  16. 16. Decisions made regarding Ho (Reject Ho/Do not reject Ho)If we reject Ho, it means it is wrong! If we do not reject Ho, it doesn’t mean it is correct, we just don’t have enough evidence to reject it! © 1984-1994 T/Maker Co.
  17. 17. Testing the Significance of Difference Between Means Z-test σ is known n ≥ 30 t-test σ is unknown n < 30 F-test 3 or more µ s (ANOVA)
  18. 18. Testing the Significance of Difference Between Means “n is large or when n ≥ 30 & σ is known.” Z-test σ is known n ≥ 30 • Hypothesized/population mean VS Sample mean and population standard deviation is known. Z= (x − µ ) n x - is the sample mean µ - is the population mean σ n - is the sample size σ - is the population std. dev. Using PHStat: Go to…“One-Sample Tests; Z-Test for the Mean:Sigma Known”
  19. 19. Testing the Significance of Difference Between Means “n is large or when n ≥ 30 & σ is known.” Z-test σ is known n ≥ 30 • Sample mean 1 VS Sample mean 2 and population standard deviation is known. x1 − x 2 x1 - is the mean of sample 1 Z= x 2 - is the mean of sample 2 1 1 σ + n1 & n2 - are the sample n1 n2 σ - is the population std. dev. sizes
  20. 20. Testing the Significance of Difference Between Means “n is large or when n ≥ 30 & σ is known.” Z-test σ is unknown n ≥ 30 • Sample mean 1 VS Sample mean 2 and 2 sample standard deviations are known. x1 - is the mean of sample 1 x1 − x 2 Z= x 2 - is the mean of sample 2 s12 s2 2 n1 & n2 - are the sample + n1 n2 s1 & s2 - are the sample std. devs. sizesUsing Microsoft Excel: Go to…“Z-Test: Two-Sample For Means”
  21. 21. The Critical Value Approach in Testing the Significance of Difference Between MeansThe 5-step solutionStep 1. Formulate Ho and HaStep 2. Set the level of significance α , usually it is given in the problem.Step 3. Formulate the decision rule (when to reject Ho); Find the critical value/P-value.Step 4. Make your decision.Step 5. Formulate your conclusion.
  22. 22. Approaches in Hypothesis Testing Critical value p- value approach approach Computed vs. Critical p-valueα vs.α α ≤5-step solution 5-step solution1.Ho: ___________ 1.Ho : ___________ Ha: ___________ Ha : ___________2.α = ___; Cri-value= ______ 2. α = ___; p- value=________ α3. Decision rule: Reject Ho 3. Decision rule: Reject Ho if if Comp − value ≥ Cri − value p- value ≤ α ≤α4. Decision: α 4. Decision:5. Conclusion: 5. Conclusion:
  23. 23. Finding Critical Values: One-Tailed What Is Z Given α = 0.05? .45 .05 Α= α = .05 Z 4 5 Z=1.65 Critical value 1.5 .4382 .4394 1.6 .4495 .4505
  24. 24. Critical Values: Z - Table α .01 0.05 Type One-T ± 2.33 ± 1.65 Two-T ± 2.58 ± 1.96You will refer to this table to get the critical value of Zor the Z tabular .
  25. 25. CRITERION:1. One-tailed test (right directional) “Reject H0 if Zc ≥ Zt “2. One-tailed test (left directional) “Reject H0 if Zc ≤ Zt3. Two-tailed test (Zc = +) “Reject H0 if Zc ≥ Zt “4. Two-tailed test (Zc = -) ‘Reject H0 if Zc ≤ Zt “
  26. 26. EXERCISES:1. Past records showed that the average final examination grade of students in Statistics was 70 with standard deviation of 8.0. A random sample of 100 students was taken and found to have a mean final examination grade of 71.8. Is this an indication that the sample grade is better than the rest of the students? Test at 0.05 level of significance.
  27. 27. 2. A certain type of battery is known to have a mean life of 60 hours. In random sample of 40 batteries, the mean life was found to be 58 hours with a standard deviation of 4.5 hours. Does it indicate that the mean lifetime of such battery has been reduced? Test at 0.01 level of significance.
  28. 28. 3. The manager of a rent-a-car business wants to know whether the true average numbers of cars rented a day is 25 with a standard deviation of 6.9 rentals. A random sample of 30 days was taken and found to have an average of 22.8 rentals. Is there a significance between the mean and the sample mean? Test at 0.05 level of significance.
  29. 29. 4. Advertisements claim that the average nicotine content of a certain kind of cigarette is 0.30 milligram. Suspecting that this figure is too low, a consumer protection service takes a random sample of 50 of these cigarette from different production slots and find that their nicotine content has a mean of 0.33 milligram with a standard deviation of 0.18 milligram. Use the 0.05 level of significance to test the null hypothesis µ = 0.30 against the alternative hypothesis µ < 0.30.
  30. 30. 5. An experiment was planned to compare the mean time (in days) required to recover from common cold for person given a daily doze of 4 mgs. of vitamin C versus those who were not given a vitamin supplement. Suppose that 35 adults were randomly selected for each treatment category and that the mean recovery times and standard deviations for the 2 groups were as follows: n X δ W/ vit. C 35 5.8 1.2 W/o vit. C 35 6.9 5.8 Suppose your research objective is to show that the use of vit. C increases the mean time required to recover from common cold. Test using α = 0.05.

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