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BCOR 1020 Business Statistics Lecture 20 – April 3, 2008

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Overview Chapter 9 – Hypothesis Testing –Problem: Testing A Hypothesis on a Proportion –Testing a Mean ( ): Population Variance ( ) Known –Problem: Testing a Mean ( ) when Population Variance ( ) is Known

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Problem: Testing A Hypothesis on a Proportion Suppose we revisited our earlier example and conducted an additional survey to determine whether the proportion of our target market that is willing to pay $25 per unit exceeds 20% (as required by the business case). In our new survey, 72 out of 300 respondents said they would be willing to pay $25 per unit. At the 5% level of significance, conduct the appropriate hypothesis test to determine whether the population proportion exceeds 20%. Include the following: –State the level of significance, . –State the null and alternative hypotheses, H 0 and H 1. –Compute the test statistic –State the decision criteria –State your decision (Overhead)

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Problem: Testing A Hypothesis on a Proportion Work: = H 0 : H 1 : Test Statistic (and Distribution under H 0 )… Decision Criteria… Decision…

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Clickers What are the appropriate null and alternative hypotheses? (B) H 0 : > H 1 : < (A) H 0 : < H 1 : > (C) H 0 : = H 1 :

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Clickers What is the point estimate of the population proportion ? (A) p = 0.15 (B) p = 0.20 (C) p = 0.22 (D) p = 0.24 (E) p = 0.36

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Clickers What is the calculated value of your test statistic? (A) Z* = 2.00 (B) Z* = 1.73 (C) Z* = 0.87 (D) Z* = 0.71

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Clickers What is your decision criteria? (A) Reject H 0 if Z* < -1.645 (B) Reject H 0 if Z* < -1.960 (C) Reject H 0 if Z* > 1.960 (D) Reject H 0 if Z* > 1.645 (E) Reject H 0 if |Z*| > 1.960

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Problem: Testing A Hypothesis on a Proportion Conclusion: Since our test statistic Z* = 1.73 > Z = 1.645, we will reject H 0 in favor of H 1 : > 0.20. Based on the data in this sample, there is statistically significant evidence that the population proportion exceeds 20%.

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Chapter 9 – Testing a Mean ( known) Hypothesis Tests on ( known): If we wish to test a hypothesis about the mean of a population when is assumed to be known, we will follow the same logic… –Specify the level of significance, (given in problem or assume 10%). –State the null and alternative hypotheses, H 0 and H 1 (based on the problem statement). –Compute the test statistic and determine its distribution under H 0. –State the decision criteria (based on the hypotheses and distribution of the test statistic under H 0 ). –State your decision.

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Chapter 9 – Testing a Mean ( known) Selection of H 0 and H 1 : Remember, the conclusion we wish to test should be stated in the alternative hypothesis. Based on the problem statement, we choose from… (i)H 0 : > 0 H 1 : < 0 (ii) H 0 : < 0 H 1 : > 0 (iii) H 0 : = 0 H 1 : 0 where 0 is the null hypothesized value of (based on the problem statement).

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Chapter 9 – Testing a Mean ( known) Test Statistic: Start with the point estimate of , Recall that for a large enough n {n > 30}, is approximately normal with and If H 0 is true, then is approximately normal with and So, the following statistic will have approximately a standard normal distribution:

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Chapter 9 – Testing a Mean ( known) Decision Criteria: Just as with tests on proportions, our decision criteria will consist of comparing our test statistic to an appropriate critical point in the standard normal distribution (the distribution of Z* under H 0 ). (i)For the hypothesis test H 0 : > 0 vs. H 1 : < 0, we will reject H 0 in favor of H 1 if Z* < – Z . (ii) For the hypothesis test H 0 : 0, we will reject H 0 in favor of H 1 if Z* > Z . (iii) For the hypothesis test H 0 : = 0 vs. H 1 : 0, we will reject H 0 in favor of H 1 if |Z*| > Z /2.

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Chapter 9 – Testing a Mean ( known) Example (original motivating example): Suppose your business is planning on bringing a new product to market. There is a business case to proceed only if –the cost of production is less than $10 per unit and –At least 20% of your target market is willing to pay $25 per unit to purchase this product. How do you determine whether or not to proceed?

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Chapter 9 – Testing a Mean ( known) Motivating Example (continued) : Assume the cost of production can be modeled as a continuous variable. –You can conduct a random sample of the manufacturing process and collect cost data. –If 40 randomly selected production runs yield and average cost of $9.00 with a standard deviation of $1.00, what can you conclude? –We will test an appropriate hypothesis to determine whether the average cost of production is less than $10.00 per unit (as required by the business case). (Overhead)

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Chapter 9 – Testing a Mean ( known) Motivating Example (continued) : Conduct the Hypothesis Test: –Specify , say = 0.05 for example. –Select appropriate hypotheses. Since we want to test that the mean is less than $10, we know that 0 = 10 and H 1 should be the “<“ inequality. So we will test (i) H 0 : > 0 vs. H 1 : < 0. –Calculate the test statistic…

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Chapter 9 – Testing a Mean ( known) Motivating Example (continued) : Conduct the Hypothesis Test: –Use the Decision criteria: (i) we will reject H 0 in favor of H 1 if Z* < – Z , where Z.05 = 1.645. –State our Decision… Since Z* = -6.32 < -Z.05 = -1.645, we reject H 0 in favor of H 1. –In “plain” language… There is statistically significant evidence that the average cost of production is less than $10 per unit.

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Chapter 9 – Testing a Mean ( known) Calculating the p-value of the test: The p-value of the test is the exact probability of a type I error based on the data collected for the test. It is a measure of the plausibility of H0. –P-value = P(Reject H0 | H0 is True) based on our data. –Formula depends on which pair of hypotheses we are testing… (i)For the hypothesis test H 0 : > 0 vs. H 1 : < 0, (ii) For the hypothesis test H 0 : 0, (iii) For the hypothesis test H 0 : = 0 vs. H 1 : 0,

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Chapter 9 – Testing a Mean ( known) Example: Let’s calculate the p-value of the test in our example… We found Z* = –6.32 Since we were testing H 0 : > 0 vs. H 1 : < 0, Interpretation: If we were to reject H 0 based on the observed data, there is approximately zero probability that we would be making a type I error. Since this is smaller than = 5%, we will reject H 0.

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Problem: Testing A Hypothesis on a Mean ( Known) In an effort to get a loan, a clothing retailer has made the claim that the average daily sales at her store exceeds $7500. Historical data suggests that the purchase amount is normally distributed with a standard deviation of = $1500. In a randomly selected sample of 24 days sales data, the average daily sales were found to be = $7900. At the 10% level of significance, conduct the appropriate hypothesis test to determine whether the data supports the retailer’s claim. Include the following: –State the level of significance, . –State the null and alternative hypotheses, H 0 and H 1. –Compute the test statistic –State the decision criteria –State your decision (Overhead)

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Problem: Testing A Hypothesis on a Mean ( Known) Work: = H 0 : H 1 : Test Statistic (and Distribution under H 0 )… Decision Criteria… Decision…

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Clickers What are the appropriate null and alternative hypotheses? (A) H 0 : > H 1 : < (B) H 0 : < H 1 : > (C) H 0 : = H 1 :

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Clickers What is the calculated value of your test statistic? (A) Z* = 0.27 (B) Z* = 1.28 (C) Z* = 1.31 (D) Z* = 1.96 (E) Z* = 5.33

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Clickers What is your decision criteria? (A) Reject H 0 if Z* < -1.645 (B) Reject H 0 if Z* < -1.282 (C) Reject H 0 if Z* > 1.282 (D) Reject H 0 if Z* > 1.645 (E) Reject H 0 if |Z*| > 1.960

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Clickers What is your decision? (A) Reject H 0 in favor of H 1 (B) Fail to reject (Accept) H 0 in favor of H 1 (C) There is not enough information

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Problem: Testing A Hypothesis on a Mean ( Known) Conclusion: Since our test statistic Z* = 1.31 is greater than Z = 1.282, we will reject H 0 in favor of H 1 : > $7500. Based on the data in this sample, there is statistically significant evidence that the average daily sales at her store exceeds $7500.

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Clickers Given our test statistics Z* = 1.31, calculate the p- value for the hypothesis test H 0 : < 7500 vs. H 1 : > 7500. (A) p-value = 0.0060 (B) p-value = 0.0951 (C) p-value = 0.9940 (D) p-value = 0.9049

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