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- What is the random variable in this problem? Gender of a customer

QA 233 PRACTICE PROBLEMS PROBABILITY, SAMPLING DISTRIBUTIONS CONFIDENCE INTERVALS & HYPOTHESIS TESTING.

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- What is the random variable in this problem? Gender of a customer

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  1. QA 233 PRACTICE PROBLEMS PROBABILITY, SAMPLING DISTRIBUTIONS CONFIDENCE INTERVALS & HYPOTHESIS TESTING These problems will give you an opportunity to practice applying the material on Probability Distributions (from Chapters 5 & 6 of the ASW textbook and Slide Sets 5 & 6), Sampling Distributions (from Chapter 7 of the ASW textbook and Slide Set 7), Confidence Intervals (from Chapter 8 of the ASW textbook and Slide Set 8), and Hypothesis Testing (from Chapter 9 of the ASW textbook and Slide Set 9).

  2. 1. Suppose that exactly three-quarters of the customers who come into your sheet music store are male. If one hundred customers come into your store today, what is the probability that at least 70% are male? - What is the random variable in this problem? Gender of a customer - On what level is this rv measured? Nominal level – we are only measuring group membership - What is the statistic of interest? The number or proportion of the sample (today’s customers) who are male - What is the question we are being asked? P(p  0.70)=?

  3. 1. Suppose that exactly three-quarters of the customers who come into your sheet music store are male. If one hundred customers come into your store today, what is the probability that at least 70% are male? - How is this statistic distributed? np = 100(0.75) 5 and np = 100(1 - 0.75) 5, so by the Central Limit Theorem (CLT) the random variable is normally distributed with an expected value of p = 0.75 and a standard deviation (also called the standard error) of

  4. f(p) 0.70 0.75 p 1. Suppose that exactly three-quarters of the customers who come into your sheet music store are male. If one hundred customers come into your store today, what is the probability that at least 70% are male? - What does this question look like? Area of Probability = 0.50 Area of Probability = ? { }

  5. 1. Suppose that exactly three-quarters of the customers who come into your sheet music store are male. If one hundred customers come into your store today, what is the probability that at least 70% are male? - How can we use this information to answer the question we are being asked? Since p (the sample proportion) is approximately normally distributed with an expected value of p = 0.75 and a standard deviation (also called the standard error) of sp = 0.043301, we can use the z-transformation to rewrite the question being asked P(p  0.70)=? in terms of the standard normal random variable z

  6. f(p) 0.70 0.75 p ???? 0.00 z 1. Suppose that exactly three-quarters of the customers who come into your sheet music store are male. If one hundred customers come into your store today, what is the probability that at least 70% are male? - What does the z-transformation look like? Area of Probability = 0.50 Area of Probability = ? { }

  7. 1. Suppose that exactly three-quarters of the customers who come into your sheet music store are male. If one hundred customers come into your store today, what is the probability that at least 70% are male? - How do we do the z-transformation?

  8. f(p) 0.70 0.75 p -1.15 0.00 z 1. Suppose that exactly three-quarters of the customers who come into your sheet music store are male. If one hundred customers come into your store today, what is the probability that at least 70% are male? - What does the z-transformation look like? Area of Probability = 0.3749 Area of Probability = 0.50 { }

  9. 2. Suppose that exactly three-quarters of the customers who come into your sheet music store are male. What is the probability that no more than one of the next six customers that come into your store today is male? - What is the random variable in this problem? Gender of a customer - On what level is this rv measured? Nominal level – we are only measuring group membership - What is the value of interest? The number of the sample (the next six customers) who are male - What is the question we are being asked? P(x  1)=?

  10. 2. Suppose that exactly three-quarters of the customers who come into your sheet music store are male. What is the probability that no more than one of the next six customers that come into your store today is male? - How is this statistic distributed? np = 6(0.75)< 5 and n(1-p) = 6(1 - 0.75)< 5so the random variable is not normally distributed – we have six trials, each with two potential outcomes, so it is binomially distributed with p = 0.75 and n = 6.

  11. 2. Suppose that exactly three-quarters of the customers who come into your sheet music store are male. What is the probability that no more than one of the next six customers that come into your store today is male? - What does this question look like? Area of Probability = ? }

  12. 2. Suppose that exactly three-quarters of the customers who come into your sheet music store are male. What is the probability that no more than one of the next six customers that come into your store today is male? - How can we use this information to answer the question we are being asked? Since x (the number of males in the next six customers) is binomially distributed with p = 0.75 and n = 6, we can calculate this probability directly as follows P(x  1) =P(x = 0) + P(x = 1) = f(0) + f(1), and

  13. and So P(x  1) = P(x = 0) + P(x = 1) = f(0) + f(1) = 0.000244+ 0.004395 = 0.004639

  14. 3. Suppose that the annual amount given by mid-level managers to the United Way fund drive is normally distributed with a mean of $135.00 and a standard deviation of 55.0. What is the probability that the next annual contribution from a mid-level manager will be for at least $160.00? - What is the random variable in this problem? Annual amount contributed by a mid-level manager to the United Way fund drive - On what level is this rv measured? Ratio level – the differences between values have meaning and there is a natural, well-defined 0 ($0.00) - What is the value of interest? The contribution made by the next mid-level manager

  15. 3. Suppose that the annual amount given by mid-level managers to the United Way fund drive is normally distributed with a mean of $135.00 and a standard deviation of 55.0. What is the probability that the next annual contribution from a mid-level manager will be for at least $160.00? - What is the question we are being asked? P(x  $160.00) = ? - How is this statistic distributed? We are told that the random variable is normally distributed with an expected value of m = $135.00 and a standard deviation of 55.00.

  16. f(x) $135.00 $160.00 x 3. Suppose that the annual amount given by mid-level managers to the United Way fund drive is normally distributed with a mean of $135.00 and a standard deviation of 55.0. What is the probability that the next annual contribution from a mid-level manager will be for at least $160.00? Area of Probability = 0.50 - What does this question look like? { Area of Probability = ? Area of Probability = ? } }

  17. 3. Suppose that the annual amount given by mid-level managers to the United Way fund drive is normally distributed with a mean of $135.00 and a standard deviation of 55.0. What is the probability that the next annual contribution from a mid-level manager will be for at least $160.00? - How can we use this information to answer the question we are being asked? Since x (the contribution made by an individual mid-level manager) is normally distributed with an expected value of m = $135.00 and a standard deviation of s = 55.0, we can use the z-transformation to rewrite the question being asked P(x  $160.00)=? in terms of the standard normal random variable z

  18. f(x) $135.00 $160.00 x 0.00 ???? z 3. Suppose that the annual amount given by mid-level managers to the United Way fund drive is normally distributed with a mean of $135.00 and a standard deviation of 55.0. What is the probability that the next annual contribution from a mid-level manager will be for at least $160.00? Area of Probability = 0.50 - What does the z-transformation look like? { Area of Probability = ? Area of Probability = ? } }

  19. 3. Suppose that the annual amount given by mid-level managers to the United Way fund drive is normally distributed with a mean of $135.00 and a standard deviation of 55.0. What is the probability that the next annual contribution from a mid-level manager will be for at least $160.00? - How do we do the z-transformation?

  20. f(x) $135.00 $160.00 x 0.00 0.45 z 3. Suppose that the annual amount given by mid-level managers to the United Way fund drive is normally distributed with a mean of $135.00 and a standard deviation of 55.0. What is the probability that the next annual contribution from a mid-level manager will be for at least $160.00? Area of Probability = 0.50 - What does the z-transformation look like? { Area of Probability = 0.1736 Area of Probability = 0.3264 } }

  21. 4. Suppose we take a sample of thirty-four mid-level managers and record the amounts they gave to the United Way fund drive last year. If the sample has a mean of $150.00 and a standard deviation of 50.0, what would be the appropriate 95% confidence interval estimate of the mean annual contribution made by mid-level managers to the United Way fund drive? - What is the random variable in this problem? Annual amount contributed by a mid-level manager to the United Way fund drive - On what level is this rv measured? Ratio level – the differences between values have meaning and there is a natural, well-defined 0 ($0.00) - What is the value of interest? The mean contribution made by mid-level managers

  22. 4. Suppose we take a sample of thirty-four mid-level managers and record the amounts they gave to the United Way fund drive last year. If the sample has a mean of $150.00 and a standard deviation of 50.0, what would be the appropriate 95% confidence interval estimate of the mean annual contribution made by mid-level managers to the United Way fund drive? - What is the question we are being asked? What symmetric interval, when calculated over infinitely many repeated samples, will contain the true mean contribution made by mid-level managers 95% of the time? - How is this statistic distributed? Since n = 34  30, the random variable is normally distributed with an expected value of m = ?? and a standard error of .

  23. 4. Suppose we take a sample of thirty-four mid-level managers and record the amounts they gave to the United Way fund drive last year. If the sample has a mean of $150.00 and a standard deviation of 50.0, what would be the appropriate 95% confidence interval estimate of the mean annual contribution made by mid-level managers to the United Way fund drive? - What is the appropriate approach to building the confidence interval? We are trying to estimate m (the population mean). Since we have a large sample (n = 34  30) the Central Limit Theorem ensures us that the distribution of sample means ( ) will be approximately normal. Also, although we don’t know the population standard deviation s, the large sample size ensures that the sample standard deviation s will be a very good estimate of s. Thus we will use

  24. f( ) 4. Suppose we take a sample of thirty-four mid-level managers and record the amounts they gave to the United Way fund drive last year. If the sample has a mean of $150.00 and a standard deviation of 50.0, what would be the appropriate 95% confidence interval estimate of the mean annual contribution made by mid-level managers to the United Way fund drive? - What is the appropriate 95% confidence interval? We have a confidence level of 95%, so z = 1.96 Area of Probability = 0.475 } $150.00 -1.96 1.96 z 0

  25. 4. Suppose we take a sample of thirty-four mid-level managers and record the amounts they gave to the United Way fund drive last year. If the sample has a mean of $150.00 and a standard deviation of 50.0, what would be the appropriate 95% confidence interval estimate of the mean annual contribution made by mid-level managers to the United Way fund drive? - We have that = $150.00, s = 50, n = 34, and z = 1.96, so the appropriate 95% confidence interval is - Interpretation: If we repeated this sampling process an infinite number of times,we would expect 95% of the resulting intervals to contain the mean contribution of all mid-level managers in the population.

  26. 5. Suppose that we believe that the mean annual contribution made by mid-level managers to the United Way fund drive is at least $165.00. Based on the sample to which we referred in the previous problem, what does a hypothesis test suggest about our conjecture at a 0.05 level of significance? At a 0.01 level of significance? 1 State the Null and Alternative Hypotheses Ho:  165.00 Ha:  < 165.00 2 Select the Appropriate Test Statistic – n = 34  30, so use

  27. f( ) 3 State the Desired Level of Significance a, Find the Critical Value(s) and State the Decision Rule First consider a=0.05. We have a one-tailed test, so za = -1.645 0.5-a=0.45 0.500 a=0.05 95% Do Not Reject Region Reject Region 0 z Decision rule – do not reject Ho if –1.645  z otherwise reject Ho

  28. 4 Calculate the Test Statistic 5 Use the Decision Rule to Evaluate the Test Statistic and Decide Whether to Reject or Not Reject the Null Hypothesis –za z i.e., –1.645  -1.74929 so reject Ho. The sample evidence does not support the claim that the mean contribution by mid-level managers is at least $165.00.

  29. We now consider a=0.01 (and repeat the hypothesis testing procedure) 1 State the Null and Alternative Hypotheses Ho:  165.00 Ha:  < 165.00 2 Select the Appropriate Test Statistic – n = 34  30, so use

  30. f( ) 4 State the Desired Level of Significance a, Find the Critical Value(s) and State the Decision Rule Since a=0.01 and we have a one-tailed test, za = -2.33 0.5-a=0.49 0.500 a=0.01 99% Do Not Reject Region Reject Region 0 z Decision rule – do not reject Ho if –2.33  z otherwise reject Ho

  31. 5 Calculate the Test Statistic 6 Use the Decision Rule to Evaluate the Test Statistic and Decide Whether to Reject or Not Reject the Null Hypothesis –za z i.e., –2.33  -1.74929 so do not reject Ho. The sample evidence supports the claim that the mean contribution by mid-level managers is at least $165.00.

  32. Notice two things about what happened when we changed the level of significance (a) from .05 to .01: - Only steps 3 and 5 of the hypothesis testing procedure were impacted - The final decision changed from reject to do not reject! This is why it is so important to perform steps 1 – 3 BEFORE you collect or analyze sample data!

  33. 6. Suppose that the annual amount given by mid-level managers to the United Way fund drive is normally distributed with a mean of $135.00 and a standard deviation of 55.0. What is the probability that the mean annual contribution of sixteen mid-level managers at SafetyCo will be no more than $150.00? - What is the random variable in this problem? Annual amount contributed by a mid-level manager to the United Way fund drive - On what level is this rv measured? Ratio level – the differences between values have meaning and there is a natural, well-defined 0 ($0.00) - What is the statistic of interest? The mean contribution made by sixteen SafetyCo mid-level managers

  34. 6. Suppose that the annual amount given by mid-level managers to the United Way fund drive is normally distributed with a mean of $135.00 and a standard deviation of 55.0. What is the probability that the mean annual contribution of sixteen mid-level managers at SafetyCo will be no more than $150.00? - What is the question we are being asked? P(  $150.00) = ?

  35. 6. Suppose that the annual amount given by mid-level managers to the United Way fund drive is normally distributed with a mean of $135.00 and a standard deviation of 55.0. What is the probability that the mean annual contribution of sixteen mid-level managers at SafetyCo will be no more than $150.00? - How is this statistic distributed? Since our parent population from which we have drawn our sample (the contribution made by an individual mid-level manager) isnormally distributed with an expected value of m = $135.00 and a standard deviation of 55.00, we have that is also normally distributed with an expected value of and a standard deviation (a.k.a. standard error) of

  36. f( ) $135.00 $150.00 6. Suppose that the annual amount given by mid-level managers to the United Way fund drive is normally distributed with a mean of $135.00 and a standard deviation of 55.0. What is the probability that the mean annual contribution of sixteen mid-level managers at SafetyCo will be no more than $150.00? - What does this question look like? Area of Probability = 0.50 { Area of Probability = ? }

  37. 6. Suppose that the annual amount given by mid-level managers to the United Way fund drive is normally distributed with a mean of $135.00 and a standard deviation of 55.0. What is the probability that the mean annual contribution of sixteen mid-level managers at SafetyCo will be no more than $150.00? - How can we use this information to answer the question we are being asked? Since (the sample mean) is normally distributed with an expected value of m = $135.00 and a standard error of = 13.75, we can use the z-transformation to rewrite the question being asked P(  $150.00) = ? in terms of the standard normal random variable z

  38. f( ) $135.00 $150.00 0.00 ???? z 6. Suppose that the annual amount given by mid-level managers to the United Way fund drive is normally distributed with a mean of $135.00 and a standard deviation of 55.0. What is the probability that the mean annual contribution of sixteen mid-level managers at SafetyCo will be no more than $150.00? - What does this question look like? Area of Probability = 0.50 { Area of Probability = ? }

  39. 6. Suppose that the annual amount given by mid-level managers to the United Way fund drive is normally distributed with a mean of $135.00 and a standard deviation of 55.0. What is the probability that the mean annual contribution of sixteen mid-level managers at SafetyCo will be no more than $150.00? - How do we do the z-transformation?

  40. f( ) $135.00 $150.00 0.00 1.09 z 6. Suppose that the annual amount given by mid-level managers to the United Way fund drive is normally distributed with a mean of $135.00 and a standard deviation of 55.0. What is the probability that the mean annual contribution of sixteen mid-level managers at SafetyCo will be no more than $150.00? - What does this question look like? Area of Probability = 0.50 { Area of Probability = 0.1736 }

  41. Notice that we did not need to use the notion of the Central Limit Theorem for Problem #4 – WHY NOT?

  42. 7. Suppose that of our sample of thirty-four mid-level managers, nineteen contributed at least $165.00 to the United Way fund drive last year. What would be the appropriate 87% confidence interval estimate of the proportion of mid-level managers that contributed at least $165.00 to the United Way fund drive last year? - What is the random variable in this problem? Whether individual mid-level managers contributed at least $165.00 to the United Way fund drive last year - On what level is this rv measured? Nominal level – the rv represents group membership (contributed at least $165.00 vs. contributed less than $165.00 to the United Way fund drive last year) - What is the value of interest? The proportion of mid-level managers who contributed at least $165.00 to the United Way fund drive last year

  43. 7. Suppose that of our sample of thirty-four mid-level managers, nineteen contributed at least $165.00 to the United Way fund drive last year. What would be the appropriate 87% confidence interval estimate of the proportion of mid-level managers that contributed at least $165.00 to the United Way fund drive last year? - What is the question we are being asked? What symmetric interval, when calculated over infinitely many repeated samples, will contain the true proportion who contributed at least $165.00 to the United Way fund drive last year? - How is this statistic distributed? Since n = 34  30, the random variable p is normally distributed with an expected value of p = ? and a standard error of

  44. 7. Suppose that of our sample of thirty-four mid-level managers, nineteen contributed at least $165.00 to the United Way fund drive last year. What would be the appropriate 87% confidence interval estimate of the proportion of mid-level managers that contributed at least $165.00 to the United Way fund drive last year? - What is the appropriate approach to building the confidence interval? We are trying to estimate p (the population proportion). Since we have a large sample (np = 34(.5)  30 and n(1 – p) = 34(.5)  30) the Central Limit Theorem ensures us that the distribution of sample proportions (p) will be approximately normal. Thus we will use

  45. 7. Suppose that of our sample of thirty-four mid-level managers, nineteen contributed at least $165.00 to the United Way fund drive last year. What would be the appropriate 87% confidence interval estimate of the proportion of mid-level managers that contributed at least $165.00 to the United Way fund drive last year? - What is the appropriate 87% confidence interval? We have a confidence level of 87%, so z = 1.51 f(p) Area of Probability = 0.435 } -1.51 1.51 z 0

  46. 7. Suppose that of our sample of thirty-four mid-level managers, nineteen contributed at least $165.00 to the United Way fund drive last year. What would be the appropriate 87% confidence interval estimate of the proportion of mid-level managers that contributed at least $165.00 to the United Way fund drive last year? - We have that p = 0.5588 and n = 34 so the standard error is

  47. 7. Suppose that of our sample of thirty-four mid-level managers, nineteen contributed at least $165.00 to the United Way fund drive last year. What would be the appropriate 87% confidence interval estimate of the proportion of mid-level managers that contributed at least $165.00 to the United Way fund drive last year? - Thus the appropriate 95% confidence interval is - Interpretation: If we repeated this sampling process an infinite number of times,we would expect 87% of the resulting intervals to contain the proportion of all mid-level managers in the population who contributed at least $165.00 to the United Way fund drive last year.

  48. 8. Suppose that we believe that no more than 35% of mid-level managers contribution at least $165.00 annually to the United Way fund drive. From our sample of thirty-four mid-level managers, nineteen actually did contribute at least $165.00 to the United Way fund drive last year. Based on the sample to which we referred in the previous problem, what hat does a hypothesis test suggest about our conjecture at a 0.05 level of significance? At a 0.01 level of significance? 1 State the Null and Alternative Hypotheses Ho: p 0.35 Ha: p > 0.35 2 Select the Appropriate Test Statistic – np0 = 11.9  5 and n(1 - p0) = 22.1  5, so use

  49. 3 State the Desired Level of Significance a, Find the Critical Value(s) and State the Decision Rule a=0.05 and we have a one (upper)-tailed test, so za = 1.65 f(z) 0.5 0.5-a=0.45 a=0.05 99% Do Not Reject Region Reject Region 0 z Decision rule – do not reject Ho if z  1.65 otherwise reject Ho

  50. 4 Calculate the Test Statistic 5 Use the Decision Rule to Evaluate the Test Statistic and Decide Whether to Reject or Not Reject the Null Hypothesis z  za i.e., 2.552867  1.65 so reject Ho. The sample evidence does not support the claim that no more than 40% of mid-level managers contribution at least $165.00 annually to the United Way fund drive.

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