1. 1 ES The Nature of Hypothesis Testing Formal process for making an inference Consider many of the concepts of a hypothesis test and look at several decision-making situations The entire process starts by identifying something of concern and then formulating two hypotheses about it

2. 2 ES Hypothesis Hypothesis: A statement that something is true Statistical Hypothesis Test: A process by which a decision is made between two opposing hypotheses. The two opposing hypotheses are formulated so that each hypothesis is the negation of the other. (That way one of them is always true, and the other one is always false). Then one hypothesis is tested in hopes that it can be shown to be a very improbable occurrence thereby implying the other hypothesis is the likely truth.

3. 3 ES Null & Alternative Hypothesis There are two hypotheses involved in making a decision: Null Hypothesis, H o : The hypothesis to be tested. Assumed to be true. Usually a statement that a population parameter has a specific value. The “starting point” for the investigation. Alternative Hypothesis, H a : A statement about the same population parameter that is used in the null hypothesis. Generally this is a statement that specifies the population parameter has a value different, in some way, from the value given in the null hypothesis. The rejection of the null hypothesis will imply the likely truth of this alternative hypothesis.

4. 4 ES Notes 1.Basic idea: proof by contradiction Assume the null hypothesis is true and look for evidence to suggest that it is false 2.Null hypothesis: the status quo A statement about a population parameter that is assumed to be true 3.Alternative hypothesis: also called the research hypothesis Generally, what you are trying to prove? We hope experimental evidence will suggest the alternative hypothesis is true by showing the unlikeliness of the truth of the null hypothesis

5. 5 ES Example Example:Suppose you are investigating the effects of a new pain reliever. You hope the new drug relieves minor muscle aches and pains longer than the leading pain reliever. State the null and alternative hypotheses. Solutions: H o :The new pain reliever is no better than the leading pain reliever H a :The new pain reliever lasts longer than the leading pain reliever

6. 6 ES Example Example:You are investigating the presence of radon in homes being built in a new development. If the mean level of radon is greater than 4 then send a warning to all home owners in the development. State the null and alternative hypotheses. Solutions: H o :The mean level of radon for homes in the development is 4 (or less) H a :The mean level of radon for homes in the development is greater than 4

7. 7 ES Null Hypothesis DecisionTrueFalse Fail to reject H o Type A correct decisionType II error RejectH o Type I errorType B correct decision Hypothesis Test Outcomes Type A correct decision: Null hypothesis true, decide in its favor Type B correct decision: Null hypothesis false, decide in favor of alternative hypothesis Type I error: Null hypothesis true, decide in favor of alternative hypothesis Type II error: Null hypothesis false, decide in favor of null hypothesis

8. 8 ES Example Example:A calculator company has just received a large shipment of parts used to make the screens on graphing calculators. They consider the shipment acceptable if the proportion of defective parts is 0.01 (or less). If the proportion of defective parts is greater than 0.01 the shipment is unacceptable and returned to the manufacturer. State the null and alternative hypotheses, and describe the four possible outcomes and the resulting actions that would occur for this test. Solutions: H o :The proportion of defective parts is 0.01 (or less) H a :The proportion of defective parts is greater than 0.01

9. 9 ES Null Hypothesis Is True: Type A correct decision Truth of situation: The proportion of defective parts is 0.01 (or less) Conclusion: It was determined that the proportion of defective parts is 0.01 (or less) Action: The calculator company received parts with an acceptable proportion of defectives Null Hypothesis Is False: Type II error Truth of situation: The proportion of defective parts is greater than 0.01 Conclusion: It was determined that the proportion of defective parts is 0.01 (or less) Action: The calculator company received parts with an unacceptable proportion of defectives Fail To Reject H o

10. 10 ES Null hypothesis is true: Type I error Truth of situation: The proportion of defectives is 0.01 (or less) Conclusion: It was determined that the proportion of defectives is greater than 0.01 Action: Send the shipment back to the manufacturer. The proportion of defectives is acceptable Null hypothesis is false: Type B correct decision Truth of situation: The proportion of defectives is greater than 0.01 Conclusion: It was determined that the proportion of defectives is greater than 0.01 Action: Send the shipment back to the manufacturer. The proportion of defectives is unacceptable Reject H o

11. 11 ES Correct DecisionTypeProbability Failure to reject a true H o A1 -  Rejection of a false H o B1 -  Errors Notes: 1.The type II error sometimes results in what represents a lost opportunity 2.Since we make a decision based on a sample, there is always the chance of making an error Probability of a type I error =  Probability of a type II error =  Error in DecisionTypeProbability Rejection of a true H o I  Failure to reject a false H o II 

12. 12 ES Notes 1.Would like  and  to be as small as possible 2.  and  are inversely related 3.Usually set  (and don’t worry too much about  Why?) 4.Most common values for  and  are 0.01 and 0.05 5.1 -  : the power of the statistical test A measure of the ability of a hypothesis test to reject a false null hypothesis 6.Regardless of the outcome of a hypothesis test, we never really know for sure if we have made the correct decision

13. 13 ES  error) I (typeP  error) II (typeP Interrelationship Interrelationship between the probability of a type I error (  ), the probability of a type II error (  ), and the sample size (n)

14. 14 ES Level of Significance & Test Statistic Level of Significance,  : The probability of committing the type I error Test Statistic: A random variable whose value is calculated from the sample data and is used in making the decision fail to reject H o or reject H o Notes: The value of the test statistic is used in conjunction with a decision rule to determine fail to reject H o or reject H o The decision rule is established prior to collecting the data and specifies how you will reach the decision

15. 15 ES The Conclusion a.If the decision is reject H o, then the conclusion should be worded something like, “There is sufficient evidence at the  level of significance to show that... (the meaning of the alternative hypothesis)” b.If the decision is fail to reject H o, then the conclusion should be worded something like, “There is not sufficient evidence at the  level of significance to show that... (the meaning of the alternative hypothesis)” Notes: The decision is about H o The conclusion is a statement about H a There is always the chance of making an error

16. 16 ES Hypothesis Test of Mean  (  known): A Probability-Value Approach The concepts and much of the reasoning behind hypothesis tests are given in the previous sections Formalize the hypothesis test procedure as it applies to statements concerning the mean  of a population (  known): a probability-value approach

17. 17 ES The Assumption... Hypothesis test: 1.A well-organized, step-by-step procedure used to make a decision 2.Probability-value approach (p-value approach): a procedure that has gained popularity in recent years. Organized into five steps. The assumption for hypothesis tests about a mean  using a known  : The sampling distribution of has a normal distribution x Recall: 1. The distribution of has mean  2. The distribution of has standard deviation x n  x

18. 18 ES The Probability-Value Hypothesis Test A Five-Step Procedure: 1.The Set-Up a.Describe the population parameter of concern b.State the null hypothesis (H o ) and the alternative hypothesis (H a ) 2.The Hypothesis Test Criteria a.Check the assumptions b.Identify the probability distribution and the test statistic formula to be used c.Determine the level of significance,  3.The Sample Evidence a.Collect the sample information b.Calculate the value of the test statistic 4.The Probability Distribution a.Calculate the p-value for the test statistic b.Determine whether or not the p-value is smaller than  5.The Results a.State the decision about H o b.State a conclusion about H a

19. 19 ES Example Example:A company advertises the net weight of its cereal is 24 ounces. A consumer group suspects the boxes are underfilled. They cannot check every box of cereal, so a sample of cereal boxes will be examined. A decision will be made about the true mean weight based on the sample mean. State the consumer group’s null and alternative hypotheses. Assume  = 0.2 Solution: 1.The Set-Up a.Describe the population parameter of concern The population parameter of interest is the mean , the mean weight of the cereal boxes

20. 20 ES b.State the null hypothesis (H o ) and the alternative hypothesis (H a ) Formulate two opposing statements concerning  H o :  = 24 ( ) (the mean is at least 24) H a :  < 24 (the mean is less than 24) Solution Continued

21. 21 ES Possible Statements of Null & Alternative Hypotheses Notes: The null hypothesis will be written with just the equal sign (a value is assigned) When equal is paired with less than or greater than, the combined symbol is written beside the null hypothesis as a reminder that all three signs have been accounted for in these two opposing statements.

22. 22 ES Example:A freezer is set to cool food to. If the temperature is higher, the food could spoil, and if the temperature is lower, the freezer is wasting energy. Random freezers are selected and tested as they come off the assembly line. The assembly line is stopped if there is any evidence to suggest improper cooling. State the null and alternative hypotheses. Example:An automobile manufacturer claims a new model gets at least 27 miles per gallon. A consumer groups disputes this claim and would like to show the mean miles per gallon is lower. State the null and alternative hypotheses. Solution: H o :  = 27 (  ) and H a :  < 27 Solution: H o :  = 10 and H a :   10 Examples

23. 23 ES )( : o  H )( : o  H )( : o  H at leastless than no less thanless than not less thanless than at mostmore than no more thanmore than not greater thangreater than isis not not different fromdifferent from same asnot same as Common Phrases & Their Negations

24. 24 ES 2.The Hypothesis Test Criteria a.Check the assumptions The weight of cereal boxes is probably mound shaped. A sample size of 40 should be sufficient for the CLT to apply. The sampling distribution of the sample mean can be expected to be normal. Example Continued: Weight of cereal boxes Recall: H o :  = 24 (  ) (at least 24) H a :  < 24 (less than 24) Example Continued b.Identify the probability distribution and the test statistic to be used To test the null hypothesis, ask how many standard deviations away from  is the sample mean n x z    * :statistictest

25. 25 ES 4.The Probability Distribution a.Calculate the p-value for the test statistic c.Determine the level of significance Let  = 0.05 3.The Sample Evidence a.Collect the sample information A random sample of 40 cereal boxes is examined 40 and 95.23  nx 5811.1 4020. 2495.23 *      n x z   b.Calculate the value of the test statistic (  = 0.2) Solution Continued

26. 26 ES Probability-Value, or p-Value: The probability that the test statistic could be the value it is or a more extreme value (in the direction of the alternative hypothesis) when the null hypothesis is true (Note: the symbol P will be used to represent the p-value, especially in algebraic situations) 0571.0 )58.1().1(*)(P   zPzPzzP P Probability-Value or p-Value

27. 27 ES Solution Continued b.Determine whether or not the p-value is smaller than  The p-value (0.0571) is greater than  (0.05) 5.The Results Decision Rule: a.If the p-value is less than or equal to the level of significance , then the decision must be to reject H o b.If the p-value is greater than the level of significance , then the decision must be to fail to reject H o a.State the decision about H o Decision about H o : Fail to reject H o b.Write a conclusion about H a There is not sufficient evidence at the 0.05 level of significance to show that the mean weight of cereal boxes is less than 24 ounces

28. 28 ES Notes If we fail to reject H o, there is no evidence to suggest the null hypothesis is false. This does not mean H o is true. The p-value is the area, under the curve of the probability distribution for the test statistic, that is more extreme than the calculated value of the test statistic. There are 3 separate cases for p-values. The direction (or sign) of the alternative hypothesis is the key.

29. 29 ES Finding p-Values 3.H a contains (Two-tailed) p-value = P(z |z*|) 1.H a contains > (Right tail) p-value = P(z > z*) 2.H a contains < (Left tail) p-value = P(z < z*)

30. 30 ES Example:The mean age of all shoppers at a local jewelry store is 37 years (with a standard deviation of 7 years). In an attempt to attract older adults with more disposable income, the store launched a new advertising campaign. Following the advertising, a random sample of 47 shoppers showed a mean age of 39.3. Is there sufficient evidence to suggest the advertising campaign has succeeded in attracting older customers? Solution: 1.The Set-Up a.Parameter of concern: the mean age, , of all shoppers b.The hypotheses: H o :  = 37 (  ) H a :  > 37 Example

31. 31 ES 2.The Hypothesis Test Criteria a.The assumptions: The distribution of the age of shoppers is unknown. However, the sample size is large enough for the CLT to apply. b.The test statistic: The test statistic will be z* c.The level of significance: none given We will find a p-value Solution Continued 3.39,47  xn 25.2 477 373.39 *      n x z   3.The Sample Evidence a.Sample information: b.Calculated test statistic:

32. 32 ES 4.The Probability Distribution a.The p-value: b.Determine whether or not the p-value is smaller than  A comparison is not possible, no  given Solution Continued 5.The Results Because the p-value is so small (P < 0.05), there is evidence to suggest the mean age of shoppers at the jewelry store is greater than 37

33. 33 ES p-Value The idea of the p-value is to express the degree of belief in the null hypothesis: 1.When the p-value is minuscule (like 0.0001), the null hypothesis would be rejected by everyone because the sample results are very unlikely for a true H o 2.When the p-value is fairly small (like 0.01), the evidence against H o is quite strong and H o will be rejected by many 3.When the p-value begins to get larger (say, 0.02 to 0.08), there is too much probability that data like the sample involved could have occurred even if H o were true, and the rejection of H o is not an easy decision 4.When the p-value gets large (like 0.15 or more), the data is not at all unlikely if the H o is true, and no one will reject H o

34. 34 ES p-Value Advantages & Disadvantage Advantages of p-value approach: 1.The results of the test procedure are expressed in terms of a continuous probability scale from 0.0 to 1.0, rather than simply on a reject or fail to reject basis 2.A p-value can be reported and the user of the information can decide on the strength of the evidence as it applies to his/her own situation 3.Computers can do all the calculations and report the p-value, thus eliminating the need for tables Disadvantage: 1.Tendency for people to put off determining the level of significance

35. 35 ES Example:The active ingredient for a drug is manufactured using fermentation. The standard process yields a mean of 26.5 grams (assume  = 3.2). A new mixing technique during fermentation is implemented. A random sample of 32 batches showed a sample mean 27.1. Is there any evidence to suggest the new mixing technique has changed the yield? Solution: 1.The Set-Up a.The parameter of interest is the mean yield of active ingredient,  b.The null and alternative hypotheses: H 0 :  = 26.5 H a :   26.5 Example

36. 36 ES 2The Hypothesis Test Criteria.a.Assumptions: A sample of size 32 is large enough to satisfy the CLT b.The test statistic: z c.The level of significance: find a p-value Solution Continued 06.1 322.3 5.261.27      n x z   3.The Sample Evidence a.From the sample: b.The calculated test statistic: 1.27,32  xn

37. 37 ES Solution Continued 4.The Probability Distribution a.The p-value: b.The p-value is large There is no  given in the statement of the problem 5.The Results Because the p-value is large (P = 0.2892), there is no evidence to suggest the new mixing technique has changed the mean yield

38. 38 ES Hypothesis Test of mean  (  known): A Classical Approach Concepts and reasoning behind hypothesis testing Formalize the hypothesis test procedure as it applies to statements concerning  of a population with known  : a classical approach

39. 39 ES The Assumption... Recall: 1.The distribution of has mean  2.The distribution of has standard deviation n  x x Hypothesis Test: A well-organized, step-by-step procedure used to make a decision. The classical approach is the hypothesis test process that has enjoyed popularity for many years. x The assumption for hypothesis tests about mean  using a known  : The sampling distribution of has a normal distribution

40. 40 ES The Classical Hypothesis Test 1.The Set-Up a.Describe the population parameter of concern b.State the null hypothesis (H o ) and the alternative hypothesis (H a ) 2.The Hypothesis Test Criteria a.Check the assumptions b.Identify the probability distribution and the test statistic to be used c.Determine the level of significance,  3.The Sample Evidence a.Collect the sample information b.Calculate the value of the test statistic 4.The Probability Distribution a.Determine the critical region(s) and critical value(s) b.Determine whether or not the calculated test statistic is in the critical region 5.The Results a.State the decision about H o b.State the conclusion about H a A Five-Step Procedure:

41. 41 ES Example: A company advertises the net weight of its cereal is 24 ounces. A consumer group suspects the boxes are underfilled. They cannot check every box of cereal, so a sample of cereal boxes will be examined. A decision will be made about the true mean weight based on the sample mean. State the consumer group’s null and alternative hypotheses. Assume  = 0.2 Solution: 1.The Set-Up a.Describe the population parameter of concern The population parameter of interest is the mean, , the mean weight of the cereal boxes Example

42. 42 ES b.State the null hypothesis (H o ) and the alternative hypothesis (H a ) Formulate two opposing statements concerning the  H o :  = 24 ( ) (the mean is at least 24) H a :  < 24 (the mean is less than 24) Solution Continued

43. 43 ES Possible Statements of Null & Alternative Hypotheses Notes: The null hypothesis will be written with just the equal sign (a value is assigned) When equal is paired with less than or greater than, the combined symbol is written beside the null hypothesis as a reminder that all three signs have been accounted for in these two opposing statements

44. 44 ES Example:An automobile manufacturer claims a new model gets at least 27 miles per gallon. A consumer groups disputes this claim and would like to show the mean miles per gallon is lower. State the null and alternative hypotheses. Solution: H o :  = 27 (  ) and H a :  < 27 Solution: H o :  = 10 and H a :   10 Examples Example:A freezer is set to cool food to. If the temperature is higher, the food could spoil, and if the temperature is lower, the freezer is wasting energy. Random freezers are selected and tested as they come off the assembly line. The assembly line is stopped if there is any evidence to suggest improper cooling. State the null and alternative hypotheses.

45. 45 ES )( : o  H )( : o  H )( : o  H at leastless than no less thanless than not less thanless than at mostmore than no more thanmore than not greater thangreater than isis not not different fromdifferent from same asnot same as Common Phrases & Their Negations

46. 46 ES Example Continued Example (continued): Weight of cereal boxes n x z     * :statistictest 2.The Hypothesis Test Criteria a.Check the assumptions The weight of cereal boxes is probably mound shaped. A sample size of 40 should be sufficient for the CLT to apply. The sampling distribution of the sample mean can be expected to be normal. Solution Continued: b.Identify the probability distribution and the test statistic to be used To test the null hypothesis, ask how many standard deviations away from  is the sample mean Recall: H o :  = 24 (>) (at least 24) H a :  < 24 (less than 24)

47. 47 ES c.Determine the level of significance Consider the four possible outcomes and their consequences Let  = 0.05 Solution Continued 3.The Sample Evidence a.Collect the sample information A random sample of 40 cereal boxes is examined b.Calculate the value of the test statistic (  = 0.2) 40 and 95.23  nx 5811.1 4020. 2495.23 *      n x z   4.The Probability Distribution a.Determine the critical region(s) and critical value(s)

48. 48 ES Critical Region & Critical Value(s) Critical Region: The set of values for the test statistic that will cause us to reject the null hypothesis. The set of values that are not in the critical region is called the noncritical region (sometimes called the acceptance region). Critical Value(s): The first or boundary value(s) of the critical region(s)

49. 49 ES Critical Region Critical Value Critical Region & Critical Value(s) Illustration:

50. 50 ES 5.The Results We need a decision rule Solution Continued 4.The Probability Distribution (Continued) b.Determine whether or not the calculated test statistic is in the critical region The calculated value of z, z* =  1.58, is in the noncritical region *

51. 51 ES Decision Rule Decision Rule: a.If the test statistic falls within the critical region, we will reject H o (the critical value is part of the critical region) b.If the test statistic is in the noncritical region, we will fail to reject H o a.State the decision about H o Decision: Fail to reject H o b.State the conclusion about H a Conclusion:There is not enough evidence at the 0.05 level of significance to show that the mean weight of cereal boxes is less than 24

52. 52 ES Notes 1.The null hypothesis specifies a particular value of a population parameter 2.The alternative hypothesis can take three forms. Each form dictates a specific location of the critical region(s) 3.For many hypothesis tests, the sign in the alternative hypothesis points in the direction in which the critical region is located 4.Significance level: 

53. 53 ES Example:The mean water pressure in the main water pipe from a town well should be kept at 56 psi. Anything less and several homes will have an insufficient supply, and anything greater could burst the pipe. Suppose the water pressure is checked at 47 random times. The sample mean is 57.1. (Assume  = 7). Is there any evidence to suggest the mean water pressure is different from 56? Use  = 0.01 Solution: 1.The Set-Up a.Describe the parameter of concern: The mean water pressure in the main pipe b.State the null and alternative hypotheses H o :  = 56 H a :   56 Example

54. 54 ES 2.The Hypothesis Test Criteria a.Check the assumptions: A sample of n = 47 is large enough for the CLT to apply b.Identify the test statistic The test statistic is z c.Determine the level of significance:  = 0.01 (given) 3.The Sample Evidence a.The sample information: b.Calculate the value of the test statistic: 47,1.57  nx 077.1 477 561.57      n x z   Solution Continued

55. 55 ES Solution Continued 4.The Probability Distribution a.Determine the critical regions and the critical values b.Determine whether or not the calculated test statistic is in the critical region The calculated value of z = 1.077, is in the noncritical region *

56. 56 ES 5.The Results a.State the decision about H o : Fail to reject H o b.State the conclusion about H a : There is no evidence to suggest the water pressure is different from 56 at the 0.01 level of significance Solution Continued

57. 57 ES Example: An elementary school principal claims students receive no more than 30 minutes of homework each night. A random sample of 36 students showed a sample mean of 36.8 minutes spent doing homework (assume  = 7.5). Is there any evidence to suggest the mean time spent on homework is greater than 30 minutes? Use  = 0.01 Solution: 1. The Set-Up The parameter of concern: , the mean time spent doing homework each night H o :  = 30 (  ) H a :  > 30 Example

58. 58 ES 2.The Hypothesis Test Criteria a.The sample size is n = 36, the CLT applies b.The test statistic is z c.The level of significance is given:  = 0.01 Solution Continued 44.5 365.7 308.36      n x z   3.The Sample Evidence

59. 59 ES 4.The Probability Distribution Solution Continued * The calculated value of z = 5.44, is in the critical region

60. 60 ES Solution Continued 5.The Results Decision: Reject H o Conclusion:There is sufficient evidence at the 0.01 level of significance to conclude the mean time spent on homework by the elementary students is more than 30 minutes Note:Suppose we took repeated sample of size 36. What would you expect to happen?

61. 61 ES 1.The t-statistic is used to complete a hypothesis test about a population mean  Hypothesis-Testing Procedure 2.The test statistic: 3.The calculated t is the number of estimated standard errors is from the hypothesized mean  4.Probability-Value or Classical Approach

62. 62 ES Example: A random sample of 25 students registering for classes showed the mean waiting time in the registration line was 22.6 minutes and the standard deviation was 8.0 minutes. Is there any evidence to support the student newspaper’s claim that registration time takes longer than 20 minutes? Use  = 0.05 and assume waiting time is approximately normal. Solution: 1.The Set-up a.Population parameter of concern: the mean waiting time spent in the registration line b.State the null and alternative hypotheses: H o :  = 20 (  ) (no longer than) H a :  > 20 (longer than) Example

63. 63 ES 2.The Hypothesis Test Criteria a.Check the assumptions: The sampled population is approximately normal b.Test statistic: t with df = n - 1 = 24 c.Level of significance:  = 0.05 Solution Continued 3.The Sample Evidence a.Sample information: b.Calculate the value of the test statistic:

64. 64 ES Using the p-Value Procedure: 4.The Probability Distribution a.The p-value: )24df with,625.1(P  tP Solution Continued Notes: If this hypothesis test is done with the aid of a computer, most likely the computer will compute the p-value for you Using Table C: read the p-value directly from the table for many situations: Using Table C: 0.05 < P < 0.10 b. The p-value is not smaller than the level of significance, 

65. 65 ES Using the Classical Procedure: 4.The Probability Distribution a.The critical value: t (24, 0.05) = 1.71 b.t is not in the critical region 5.The Results a.Decision: Fail to reject H o b.Conclusion:There is insufficient evidence to show the mean waiting time is greater than 20 minutes at the 0.05 level of significance Solution Continued

66. 66 ES Example: A new study indicates that higher than normal (220) cholesterol levels are a good indicator of possible heart attacks. A random sample of 27 heart attack victims showed a mean cholesterol level of 231 and a standard deviation of 20. Is there any evidence to suggest the mean cholesterol level is higher than normal for heart attack victims? Use  = 0.01 Solution: 1.The Set-up a.Population parameter of concern: The mean cholesterol level of heart attack victims b.State the the null and alternative hypothesis: H o :  = 220 (  ) (mean is not greater than 220) H a :  > 220 (mean is greater than 220) Example

67. 67 ES 2.The Hypothesis Test Criteria a.Assumptions: We will assume cholesterol level is at least approximately normal. b.Test statistic: t (  unknown), df = n  1 = 26 c.Level of significance:  = 0.01 (given) Solution Continued 3.The Sample Evidence a.Sample information: b.Calculate the value of the test statistic:

68. 68 ES 4.The Probability Distribution a.The critical value: t (26, 0.01) = 2.48 Solution Continued 5.The Results a.Decision: Reject H 0 b.Conclusion:At the 0.01 level of significance, there is sufficient evidence to suggest the mean cholesterol level in heart attack victims is higher than normal t (26, 0.01) * b.t* falls in the critical region

69. 69 ES Inferences About the Probability of Success & Proportion Possibly the most common inference of all Many examples of situations in which we are concerned about something either happening or not happening Two possible outcomes, and multiple independent trials

70. 70 ES 1.p: the binomial parameter, the probability of success on a single trial Background 2. : the observed or sample binomial probability 'p x represents the number of successes that occur in a sample consisting of n trials 3.For the binomial random variable x: 4.The distribution of x is approximately normal if n is larger than 20 and if np and nq are both larger than 5

71. 71 ES 1.a mean equal to p, 2.a standard error equal to, and 3.an approximately normal distribution if n is sufficiently large 'p  'p  Sampling Distribution of p': If a sample of size n is randomly selected from a large population with p = P(success), then the sampling distribution of p' has: Sampling Distribution of p' In practice, use of the following guidelines will ensure normality: 1.The sample size is greater than 20 2.The sample consists of less than 10% of the population 3.The products np and nq are both larger than 5

72. 72 ES The assumptions for inferences about the binomial parameter p: The n random observations forming the sample are selected independently from a population that is not changing during the sampling The Assumptions... Confidence Interval Procedure: The unbiased sample statistic p' is used to estimate the population proportion p The formula for the  confidence interval for p is: n qp p n qp z (  p '' ' to '' '  z ( 

73. 73 ES Example: A recent survey of 300 randomly selected fourth graders showed 210 participate in at least one organized sport during one calendar year. Find a 95% confidence interval for the proportion of fourth graders who participate in an organized sport during the year. Solution: 1.Describe the population parameter of concern The parameter of interest is the proportion of fourth graders who participate in an organized sport during the year Example 2.Specify the confidence interval criteria a.Check the assumptions The sample was randomly selected Each subject’s response was independent

74. 74 ES b.Identify the probability distribution z is the test statistic p' is approximately normal c.Determine the level of confidence:  0.95 Solution Continued 3.Collect and present sample evidence Sample information: n = 300, and x = 210 The point estimate:

75. 75 ES 4.Determine the confidence interval a.Determine the confidence coefficients: Using Table 4, Appendix B: z (  /2) = z (0.025) = 1.96 Solution Continued c.Find the lower and upper confidence limits: 0519.0)0265.0)(96.1(0007.0)96.1( 300 )30.0)(70.0( 96.1 ''   n qp E b.The maximum error of estimate: z (  /2)

76. 76 ES d.The Results 0.6481 to 0.7519 is a 95% confidence interval for the true proportion of fourth graders who participate in an organized sport during the year Solution Continued E: maximum error of estimate  : confidence level p*: provisional value of p (q* = 1  p*) If no provisional values for p and q are given use p* = q* = 0.5 (Always round up) 2 2 **][ E qp n   Sample Size Determination: z (  /2)

77. 77 ES Hypothesis-Testing Procedure: For hypothesis tests concerning the binomial parameter p, use the test statistic z*: Example: (Probability-Value Approach) A hospital administrator believes that at least 75% of all adults have a routine physical once every two years. A random sample of 250 adults showed 172 had physicals within the last two years. Is there any evidence to refute the administrator's claim? Use  = 0.05 Hypothesis-Testing Procedure

78. 78 ES Solution 1.The Set-up a.Population parameter of concern: the proportion of adults who have a physical every two years b.State the null and alternative hypotheses: H o : p = 0.75 (>) H a : p < 0.75 2.The Hypothesis Test Criteria a.Assumptions: 250 adults independently surveyed b.Test statistic: z n = 250 np = (250)(0.75) = 187.5 > 5 nq = (250)(0.25) = 62.5 > 5 c.Level of significance:  = 0.05

79. 79 ES Solution Continued 3.The Sample Evidence a.Sample information: b.The test statistic: 4.The Probability Distribution a.The p-value: Use Table A or use a computer

80. 80 ES 0119.0)26.2(  zPP p-value Solution Continued b.The p-value is smaller than the level of significance,  5.The Results a.Decision: Reject H o b.Conclusion:There is evidence to suggest the proportion of adults who have a routine physical exam every two years is less than 0.75 at the 0.05 level of significance

81. 81 ES Example:(Classical Procedure) A university bookstore employee in charge of ordering texts believes 65% of all students sell their statistics books back to the bookstore at the end of the class. To test this claim, 200 statistics students are selected at random and 141 plan to sell their texts back to the bookstore. Is there any evidence to suggest the proportion is different from 0.65? Use  = 0.01 Solution: 1.The Set-up a.Population parameter of concern: p = the proportion of students who sell their statistics books back to the bookstore b.The null and alternative hypotheses: H o : p = 0.65 H a : Example

82. 82 ES Solution Continued 2.The Hypothesis Test Criteria a.Assumptions: Sample randomly selected. Each subject’s response was independent of other responses. b.Test statistic: z n = 200 np = (200)(0.65) = 130 > 5 ; nq = (200)(0.35) = 70 > 5 c.Level of significance:  = 0.01 3.The Sample Evidence a. b.Calculate the value of the test statistic:

83. 83 ES 4.The Probability Distribution a.The critical value: z (0.005) = 2.58 Solution Continued 5.The Results a.Decision: Do not reject H o b.Conclusion:There is no evidence to suggest the true proportion of students who sell their statistics texts back to the bookstore is different from 0.65 at the 0.05 level of significance b.z is not in the critical region *

84. 84 ES Notes 1.There is a relationship between confidence intervals and two- tailed hypothesis tests when the level of confidence and the level of significance add up to 1 2.The confidence interval and the width of the noncritical region are the same 3.The point estimate is the center of the confidence interval, and the hypothesized mean is the center of the noncritical region 4.If the hypothesized value of p is contained in the confidence interval, then the test statistic will be in the noncritical region 5.If the hypothesized value of p does not fall within the confidence interval, then the test statistic will be in the critical region