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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
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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.
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Hypotheses Testingis deciding between what is REALITY and what is COINCIDENCE!
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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.
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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)
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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
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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
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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
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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 .
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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
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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
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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?
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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
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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)
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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”
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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
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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”
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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.
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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:
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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
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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 .
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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 “
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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.
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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.
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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.
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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.
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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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