1. How confident can we be in our analysis?

2. Unit Plan – 10 lessons  Recap on CLT and Normal Distribution  Confidence intervals for the mean  Confidence intervals for proportions  Confidence intervals for the difference between two means  Sample size  Margin of error  Proofs

3. I Can Do…

4. Starter lesson 1 : A sample of 16 items is taken from a population with  = 34 and  = 4.  Find the mean and standard deviation of the distribution of the total. E(T) = n  T = = 16 (34) = 4 (4) = 544 = 16  Find the mean and the standard deviation of the sample means. E(X) =   X = = 34 = 1  Assume population is normal!

5. This lesson aims to be a revision period for: The Normal Distribution The Central Limit Theorem Lesson One: Re-cap: The Normal Distribution and CLT

6. Distribution of the Sample Mean  When several different samples are taken from a population, the results will vary from sample to sample.  These results will have a distribution of their own  We covered these distributions in the previous unit

7. Distribution of the Sample Mean  If n (the sample size) is large enough (>30) the distribution will be approximately normal.  The distribution of the sample mean (X) has a mean of it’s own called Mu (μ) and a standard deviation of This is also known as the standard error of the sample mean. (It gives an indication of it’s spread) We will revisit this in another lesson.

8. Remember these? A sample of 16 items is taken from a population with  = 34 and  = 4.  Find the mean and standard deviation of the distribution of the total. E(T) = n  T = = 16 (34) = 4 (4) = 544 = 16  Find the mean and the standard deviation of the sample means. E(X) =   X = = 34 = 1  Assume population is normal!

9. Sample Stats and Population Parameters Sample Statistics Populations Parameters Mean = μμ Standard Deviation s =σ

10. Re-cap: The Central Limit Theorem If samples of sample size n > 30 are repeatedly taken from any population, of no matter what distribution, then X, the random variable obtained by finding the means of these samples, is approximately normally distributed with mean  and standard deviation [  is the mean and  the standard deviation of the population]

11. Applications of the Central Limit Theorem: probability questions for the distributions of the sample mean and the total  Probability questions on the sample mean may be given now that we know that this distribution (of the sample means) is approximately normally distributed. We use normal distribution to solve.  If the underlying population is normally distributed then we may also be given probability questions on the distribution of a sample total. Again, we use normal distribution to solve.

12. When is the Assumption for normality not required?  Assumption for normality is not required as the Central Limit Theorem states that for n > 30 the distribution of the sample mean is normal regardless of the underlying population distribution.

13. By the end of this lesson students should be able to: Explain how a confidence interval gives an estimate of the population mean. Calculate a confidence interval for the population parameter μ and interpret it. Lesson Two: Introduction to Confidence Intervals

14. Confidence Intervals  We use samples as estimates for the population parameters.  As different samples produce different estimates we give an interval rather than specific values as an estimate of the population mean. This interval is called a confidence interval.

15. 95% Confidence Interval  A 95% confidence interval means that on average, 95% of the time the interval will contain the true population mean μ.

16. 95% Confidence Interval  A 95% CI maybe worked out using:  The actual formula is: If it is known, otherwise use ‘s’. Size of the sample. Z value for 95%

17. Bolts  A machine manufactures bolts to a set length with a standard deviation of 2.5mm.  A random sample of 20 bolts is checked and found to have a mean length of 75.2mm.  Find the 95% confidence interval for the mean length of bolts.

18. Maths Test Marks  The population of the marks resembles a normal distribution. XiXi MiMi FiFi 20-25| 30-35|||| 40-45|||| 50-55|||| |||| |||| 60-65|||| |||| |||| | 70-75|||| || 80-85|||| 90-10095|||| Population Parameters: μ = 64.84% σ = 16.64% n = 61 Sample Statistics: μ x = _____% n = 16

19. Back to Maths Test Marks Population Parameter: μ = 64.84%

20. Ice Cream Factory

21. By the end of this lesson students should be able to: Use a sample proportion to calculate a C.I. in order to estimate π given the formula: Lesson Three: C.I. for the Population Proportion (π)

22. Definition Check Point Confidence Level Inverse (Normal Value) Sample Proportion Sample Size Population Proportion (most cases this is unknown so we would have to use p as an estimate)

23. Defective Cubes  We have a company that produces blue counters for commercial use.  We conduct regular quality control tests of our product.  Occasionally the odd yellow counter is produced much to our annoyance!

24. Defective Cubes  Collect seven (7) quality control tests. Record the proportion of yellow counters: 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0

25. Defective Cubes  Calculate the 95% confidence level for the population of yellow counters: 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0

26. 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 Defective Cubes  Here’s our data for the seven (7) quality control tests: How would this data change if we knew that the actual population proportion of yellow beans was 0.25?

27. Cleansville  A sample of 500 Cleansville residents found that 42% of them recycle their rubbish. Calculate a 99% confidence interval for the proportion of residents who recycle in Cleansville.

28. Your Turn…  Have a go at the questions on page 86 and 87 of the W.O.N. or  Exercise 14.3 Page 232 Sigma Text Book

29. By the end of this lesson students should be able to: Calculate the sample size needed for a given C.I. width for a mean and a proportion. Understand what the margin of error shows. Lesson Four: Sample Size (n)

30. The Margin of Error  The Margin of Error is half the length of the confidence interval. It is the distance between the sample mean x and one of the end points of the interval. Lowerupper Margin of error

31. The Margin of Error  To calculate the sample size, we use: e is the accuracy of the margin of error (the “width”)

32. Lets give it a go… 1. A Golf ball has a bounce which is normally distributed and σ = 3.6cm. If a sample of 100 balls is tested and x = 82cm, find a 95% C.I. 2. Find the total width of the C.I. 3. What sample size would be needed if a C.I. was within 0.5 cm with a 95% confidence?

33. Qu. 1  A Golf ball has a bounce which is normally distributed and σ = 3.6cm. If a sample of 100 balls is tested and x = 82cm, find a 95% C.I. 81.2982.71 81.29 < μ < 82.71

34. Qu. 1  Find the total width of the C.I. 81.2982.71 = 82.71-81.29 = 1.42

35. The Margin of Error  The Margin of Error is half the length of the confidence interval. It is the distance between the sample mean x and one of the end points of the interval. 81.2982.71 Margin of error (or accuracy) = 0.71

36. Cont’d…  What sample size would be needed if a C.I. was within 0.5 cm with a 95% confidence?

37. Cont’d…  What sample size would be needed if the total length of a C.I. was 1.2 cm with a 95% confidence?

38. Lesson Five: Sample size for proportions:Two Cases  There are two cases for proportion:  π is given  π is unknown

39. Two Cases – π is Given  Research has shown that 42% of households have SkyTV.  How large a sample is needed to have a 99% confidence that the sample proportion is within 4% of the true percentage?

40. Two Cases – π is Unknown  An opinion poll is to be conducted. What is the minimum size needed so that the margin of error is no greater than 3% for a 95% C.I.  Use π = 0.5 (gives largest C.I.)

41. Your Turn  W.O.N. Pages 94-96 or  Sigma Text Ex 14.4 Page 235

42. Starter lesson 6 : 1). A random sample from a population with standard deviation 14 yielded a 99% confidence interval for the mean between 72 and 84. Find the sample mean. Find the sample size. 2). A gardener wants to estimate (within 3 days) the average number of days it takes for tomatoes to grow. He knows SD of growing times is about 5 days. What is required sample size at 95% confidence level?

43. Starter solutions: 1). A random sample from a population with standard deviation 14 yielded a 99% confidence interval for the mean between 72 and 84. Find the sample mean. 72 <  < 84Mid-point is Find the sample size. First, e = 84 – 78 = 6. Now: For 99% CI: z = 2.576. So sample size is 37. So:

44. Starter solutions: 2). A gardener wants to estimate (within 3 days) the average number of days it takes for tomatoes to grow. He knows SD of growing times is about 5 days. What is required sample size at 95% confidence level? So minimum sample size is 11. 95% CI, so z = 1.96 e = 3 days  = 5 days

45. By the end of this lesson students should be able to: Calculate the difference of two means needed when comparing two populations. Lesson Six: Difference of Two Means

46. C.I. for the difference between two means  A frequent problem in statistics is to determine whether two populations are:  Similar, or  Whether there is a significant difference (and unlikely to be the same)

47. Who Drives Faster?  In a recent trial, 16 girls and 12 boys took part in a drag strip simulator to refute/confirm this statement: “Girls can drive faster than Boys”  The results were as follows…

48. Who Drives Faster? FemalesMales  x F = 165.7 km/h  S F = 45  N = 16  x M = 160 km/h  S F = 25  N = 12 Is there or is there not a significant difference between these? Calculate a 95% confidence level for both of these and graph it.

49. Who Drives Faster? Boys Girls

50. Who Drives Faster?  How do we decide whether small differences are significant or not? We have a small unreliable sample (if we are testing the idea that the population of girls can drive faster than boys.)  We need a formula to test this properly…

51. Distribution of the Difference of Two Means  We are estimating the difference between two population means (a parameter) μ 1 – μ 2  The logical statistic to use is x 1 – x 2

52. The C.I. for the Difference  A confidence interval for a parameter usually takes the form: so… (Sample value) ± (Confidence Level) × (SD of sample value)

53. Who Drives Faster?  Use the above formula to calculate the difference between the sample means (95% CI)

54. Who Drives Faster?  If there is no underlying difference between the speeds that girls and boys can drive at then: μ F – μ M = 0 (Zero)  So all we have to do is check whether zero is included in the 95% C.I.  If such a confidence interval does not enclose zero then it is unlikely that the two means are equal. There is probably a difference between the two means.

55. Who Drives Faster?  Our Conclusion:  Since _______ lies within the confidence interval, there is insufficient evidence to conclude that _________________________. (on a drag strip simulator)

56. Significant Differences  It IS possible for even smaller differences to be significant if the sample is large enough.  Consider… FemalesMales  x F = 165.7 km/h  S F = 45  N = 20 000  x M = 163.2 km/h  S F = 25  N = 18500

57. You could be asked questions on:  Constructing these CI’s  Finding margins of error for these CI’s  Interpreting these CI’s Note on these confidence intervals:

58. Formulae to Remember SituationStandard Error Individual Continuous Data Distribution of Sample Means Distribution of Sample Proportions Distribution of Difference of Two Means

59. Your Height Please…  You need to record your heights to the nearest centimetre. (Here)Here  We will need to keep males heights and females heights separate.  Calculate μ and σ (Population Parameters) for males and females Sample Statistics Populations Parameters Mean xμ Standard Deviation sσ

60. Extension: C.I. for the difference between two proportions  By the end of this lesson students should be able to: Calculate and interpret confidence intervals for the difference between two proportions

61. Extension: C.I. for the difference between two proportions  This is similar to finding the difference between two means  We estimate the population parameter π with the sample statistic p.

62. Starter lesson 7: Two independent populations have means of 85.4 and 64.3 respectively and standard deviations are 8.7 and 6.4. A random sample of 64 is drawn from the first population and 36 from the second.  Find expected value of difference between two sample means. E( ) =     = 85.4 – 64.3 = 21.1  Find SD of difference between the two sample means. =  What is the probability that the mean of the difference between these sample means will be greater than 22? Let D = difference = So P(D > 22)   = 85.4   = 64.3   = 8.7   = 6.4 n 1 = 64 n 2 = 36

63. A sample of size 50 is taken from a population X with mean 78 and a SD of 4. A sample of size 40 is taken from a population Y with mean 76 and SD of 5. Find:  E(X)  E(Y)  var(X)  var(Y)  E(X – Y)  var(X – Y)  SD(X – Y) Starter lesson 7 :

64. A sample of size 50 is taken from a population X with mean 78 and a SD of 4. A sample of size 40 is taken from a population Y with mean 76 and SD of 5. Find:  E(X) =  X = 78  E(Y) =  Y = 76  var(X) =  var(Y) =  E(X – Y) =  X  Y = 78 – 76 = 2  var(X – Y) = var(X) + var(Y) = 0.32 + 0.625 = 0.945  SD(X – Y) Starter solutions:

65. Lesson Seven: Practice Q’s  By the end of this lesson students should be able to: Calculate and interpret confidence intervals for: Means Proportions Difference between means

66. Lesson Nine: Excellence (1)  By the end of this lesson students should be able to: Find standard errors of estimates More revision

67. Lesson Ten: Excellence (2)  By the end of this lesson students should be able to: Prove the formula for the standard error of the difference between two means Understand in depth applications of the Central Limit Theorem

68. Proof of Diff.

70. Proof of Diff. Full proof

71. Lesson Ten: Practice Assessment