1. Chapter 10 Hypothesis Tests Using a Single Sample

19. Error No Error No Error Type I Error Type II Error  

27. Large Sample Hypothesis Test for a Single Proportion In terms of a standard normal random variable z, the approximate P-value for this test depends on the alternate hypothesis and is given for each of the possible alternate hypotheses on the next 3 slides. To test the hypothesis H 0 :  = hypothesized proportion, compute the z statistic

28. Hypothesis Test Large Sample Test of Population Proportion

29. Hypothesis Test Large Sample Test of Population Proportion

30. Hypothesis Test Large Sample Test of Population Proportion

32. Hypothesis Test Example 2 Single Proportion continued  = proportion of the company’s claims that are settled within 30 days H 0 :  = 0.9 H A :   0.9 The sample proportion is

33. Hypothesis Test Example 2 Single Proportion continued The probability of getting a result as strongly or more strongly in favor of the consumer group's claim (the alternate hypothesis H a ) if the company’s claim (H 0 ) was true is essentially 0. Clearly, this gives strong evidence in support of the alternate hypothesis (against the null hypothesis).

34. Hypothesis Test Example 2 Single Proportion continued We would say that we have strong support for the claim that the proportion of the insurance company’s claims that are settled within 30 days is less than 0.9. Some people would state that we have shown that the true proportion of the insurance company’s claims that are settled within 30 days is statistically significantly less than 0.9.

36. Hypothesis Test Example Single Proportion continued  = proportion of his party that is in opposition H 0 :  = 0.25 H A :  > 0.25  = 0.10 Note: hypothesized value = 0.25

40. Hypothesis Test (Large samples) Single Sample Test of Population Mean In terms of a standard normal random variable z, the approximate P-value for this test depends on the alternate hypothesis and is given for each of the possible alternate hypotheses on the next 3 slides. To test the hypothesis H 0 : µ  = hypothesized mean, compute the z statistic

41. Hypothesis Test Single Sample Test of Population Mean H 0 : µ = hypothesized mean H A : µ < hypothesized mean

42. Hypothesis Test Single Sample Test of Population Mean H 0 : µ = hypothesized mean H A : µ > hypothesized mean

43. Hypothesis Test Single Sample Test of Population Mean H 0 : µ = hypothesized mean H A : µ ≠ hypothesized mean

44. Reality Check For large values of n (>30) it is generally acceptable to use s to estimate  , however, it is much more common to apply the t-distribution. It is not likely that one would know  but not know  , so calculating a z value using the formula would not be very realistic.

45. Hypothesis Test (  unknown) Single Sample Test of Population Mean The approximate P-value for this test is found using a t random variable with degrees of freedom df = n-1. The procedure is described in the next group of slides. To test the null hypothesis µ = hypothesized mean , when we may assume that the underlying distribution is normal or approximately normal, compute the t statistic

46. Hypothesis Test Single Sample Test of Population Mean H 0 : µ = hypothesized mean H A : µ < hypothesized mean

47. Hypothesis Test Single Sample Test of Population Mean H 0 : µ = hypothesized mean H A : µ > hypothesized mean

48. Hypothesis Test Single Sample Test of Population Mean H 0 : µ = hypothesized mean H A : µ ≠ hypothesized mean

49. Hypothesis Test (  unknown) Single Sample Test of Population Mean The t statistic can be used for all sample sizes, however, the smaller the sample, the more important the assumption that the underlying distribution is normal. Typically, when n >15 the underlying distribution need only be centrally weighted and may be somewhat skewed.

50. Tail areas for t curves

51. Tail areas for t curves

54. Example of Hypothesis Test Single Sample Test of Population Mean continued

55. Example of Hypothesis Test Single Sample Test of Population Mean conclusion Because P-value = 0.087 > 0.05 =  we fail to reject H 0. At a level of significance of 0.05, there is insufficient evidence to conclude that the mean shearing strength of this brand of bolt exceeds 110 lbs.

56. Using the t table t = 1.4 n = 25 df = 24 Tail area = 0.087

62. Another Example We can see that with the exception of one outlier, the data is reasonably symmetric and mound shaped in shape, indicating that the assumption that the population of amounts of gold for this particular charm can reasonably be expected to be normally distributed.

69. Example (based on z-curve)

70. Example (based on z-curve)