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What single samples t test (2)?

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What is a Single Sample T?

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What single samples t test (2)?

  1. 1. Single Sample t-Tests
  2. 2. Welcome to a presentation explaining the concepts behind the use of a single sample t-test
  3. 3. Welcome to a presentation explaining the concepts behind the use of a single sample t-test in determining the probability that a sample and a population are similar to or different from one another statistically.
  4. 4. We will follow an example where researchers attempt to determine if the sample they have collected is statistically significantly similar or different from a population.
  5. 5. Their hope is that the sample and population are statistically similar to one another, so they can claim that results of experiments done to the sample are generalizable to the population.
  6. 6. Let’s imagine that this is the population distribution for IQ scores in the country:
  7. 7. Let’s imagine that this is the population distribution for IQ scores in the country:
  8. 8. It has a population mean of 100
  9. 9. m = 100
  10. 10. m = 100 This Greek symbol represents the mean of a population
  11. 11. We decide to select a random sample to do experiments on. m = 100
  12. 12. So, we randomly select 20 persons m = 100
  13. 13. Let’s say that sample of 20 has an IQ score mean of 70 m = 100
  14. 14. Let’s say that sample of 20 has an IQ score 풙 = 70 m = 100 mean of 70
  15. 15. 풙 = 70 m = 100 Note, this x with a bar over it is the symbol for a sample mean.
  16. 16. 풙 = 70 m = 100 Again this Greek symbol m is the symbol for a population mean.
  17. 17. Along with a mean of 70 this sample has a distribution that looks 풙 = 70 m = 100 like this
  18. 18. Along with a mean of 70 this sample has a distribution that looks 풙 = 70 m = 100 like this
  19. 19. 풙 = 70 m = 100 So, here’s the question:
  20. 20. 풙 = 70 m = 100 Is this randomly selected sample of 20 IQ scores representative of the population?
  21. 21. The Single Sample t-test is a tool used to determine the probability that it is or is not. 풙 = 70 m = 100
  22. 22. So, how do we determine if the sample is a good representative of the population?
  23. 23. Let’s look at the population distribution of IQ scores first:
  24. 24. Let’s look at the population distribution of IQ scores first:
  25. 25. One thing we notice right off is that it has a normal distribution
  26. 26. Normal Distributions have some important properties or attributes that make it possible to consider rare or common occurrences
  27. 27. Also, Normal Distributions have some constant percentages that are true across all normal distributions
  28. 28. Attribute #1: 50% of all of the scores are above the mean and the other 50% of the scores are below the mean
  29. 29. Attribute #1: 50% of all of the scores are above the mean and the other 50% of the scores are below the mean The Mean
  30. 30. Attribute #1: 50% of all of the scores are above the mean and the other 50% of the scores are below the mean The Mean 50% of all scores
  31. 31. Attribute #1: 50% of all of the scores are above the mean and the other 50% of the scores are below the mean 50% of all scores The Mean
  32. 32. Before going on let’s take a brief time out
  33. 33. The next section requires an understanding of the concept of standard deviation.
  34. 34. If you are unfamiliar with this concept do a search for standard deviation in this software. After viewing it return to slide 36 of this presentation.
  35. 35. Time in – let’s get back to the instruction
  36. 36. The Mean Attribute #2: 68% of all of the scores are between +1 standard deviation and -1 standard deviation
  37. 37. The Mean Attribute #2: 68% of all of the scores are between +1 standard deviation and -1 standard deviation from the mean
  38. 38. The Mean Attribute #2: 68% of all of the scores are between +1 standard deviation and -1 standard deviation from the mean +1 sd
  39. 39. The Mean Attribute #2: 68% of all of the scores are between +1 standard deviation and -1 standard deviation from the mean -1 sd +1 sd
  40. 40. 68% of all scores The Mean Attribute #2: 68% of all of the scores are between +1 standard deviation and -1 standard deviation from the mean -1 sd +1 sd
  41. 41. 68% of all scores The Mean So what this means is – if you were randomly selecting samples from this population you have a 68% chance or .68 probability of pulling that sample from this part of the distribution. -1 sd +1 sd
  42. 42. 68% of all scores The Mean -1 sd +1 sd Let’s put some numbers to this idea.
  43. 43. 68% of all scores The Mean -1 sd +1 sd The mean of IQ scores across the population is 100
  44. 44. -1 sd +1 sd m = 100 With a population standard deviation (s) of 15: +1 standard deviation would at an IQ score of 115 and -1 standard deviation would be at 85 68% of all scores
  45. 45. m = 100 +1 sd -1s=85 -1 sd 68% of all scores With a population standard deviation (s) of 15: +1 standard deviation would at an IQ score of 115 and -1 standard deviation would be at 85
  46. 46. 68% of all scores -1s=85 m = 100 +1s=115 -1 sd +1 sd With a population standard deviation (s) of 15: +1 standard deviation would at an IQ score of 115 and -1 standard deviation would be at 85
  47. 47. 68% of all scores So, there is a 68% m = 100 -1 sd +1 sd chance or .68 probability that a sample was collected between IQ scores of 85 and 115 -1s=85 +1s=115
  48. 48. 68% of all scores m = 100 -1 sd +1 sd Attribute 2: 2 standard deviation units above and below the mean constitute 95% of all scores. -1s=85 +1s=115
  49. 49. 68% of all scores m = 100 -1 sd +1 sd Attribute 2: 2 standard deviation units above and below the mean constitute 95% of all scores. -1s=85 +1s=115 +2 sd +2 sd
  50. 50. 68% of all scores m = 100 -1 sd +1 sd 2 standard deviation units above the mean would be an IQ score of 130 or 100 + 2*15(sd)) -1s=85 +1s=115 +2 sd +2 sd
  51. 51. 68% of all scores m = 100 -1 sd +1 sd 2 standard deviation units above the mean would be an IQ score of 130 or 100 + 2*15(sd)) -1s=85 +1s=115 +1s=130 +2 sd -2 sd
  52. 52. 68% of all scores m = 100 -1 sd +1 sd 2 standard deviation units below the mean would be an IQ score of 70 or 100 - 2*15(sd)) -1s=85 +1s=115 +1s=115 +2 sd -2 sd
  53. 53. 68% of all scores -2s=70 m = 100 +1s=115 -1 sd +1 sd 2 standard deviation units below the mean would be an IQ score of 70 or 100 - 2*15(sd)) -1s=85 +1s=115 +2 sd -2 sd
  54. 54. 68% of all scores -2s=70 m = 100 +1s=115 -1 sd +1 sd Now, it just so happens in nature that 95% of all scores are between +2 and -2 standard deviations in a normal distribution. -1s=85 +1s=115 +2 sd -2 sd
  55. 55. 95% of all scores -2s=70 -1s=85 m = 100 +1s=115 +1s=130 -1 sd +1 sd +2 sd -2 sd Now, it just so happens in nature that 95% of all scores are between +2 and -2 standard deviations in a normal distribution.
  56. 56. 95% of all scores -2s=70 m = 100 +1s=130 -1 sd +1 sd This means that there is a .95 chance that a sample we select would come from between these two points. -1s=85 +1s=115 +2 sd -2 sd
  57. 57. Attribute #3: 99% of all scores are between 95% of all scores -2s=70 -1s=85 m = 100 +1s=115 +1s=130 -1 sd +1 sd +2 sd -2 sd +3 and -3 standard deviations.
  58. 58. 99% of all scores -2s=70 -1s=85 m = 100 +1s=115 +1s=130 -1 sd +1 sd +2 sd -2 sd -3s=55 -3 sd +3s=55 +3 sd Attribute #3: 99% of all scores are between +3 and -3 standard deviations.
  59. 59. 99% of all scores -2s=70 -1s=85 m = 100 +1s=115 +1s=130 -1 sd +1 sd +2 sd -2 sd -3s=55 -3 sd -3s=55 +3 sd These standard deviations are only approximates.
  60. 60. 99% of all scores -2s=70 -1s=85 m = 100 +1s=115 +1s=130 -3s=55 -3s=55 -3 sd +3 sd -1 sd +1 sd +2 sd -2 sd Here are the actual values
  61. 61. 99% of all scores -2s=70 -1s=85 m = 100 +1s=115 +1s=130 -1 sd +1 sd +2 sd -2 sd -2.58 s =55 -2.58 sd -3s=55 +3 sd Here are the actual values
  62. 62. 99% of all scores -1.96 s -1s=85 m = 100 +1s=115 +1s=130 =70 -1 sd +1 sd +2 sd -2.58 sd -1.96 sd -3s=55 +3 sd Here are the actual values -2.58 s =55
  63. 63. 99% of all scores m = 100 -1 sd +1 sd +1s=130 +2 sd -1.96 s =70 -2.58 sd -1.96 sd -3s=55 +3 sd Here are the actual values -2.58 s =55 +1s=115 -1 s =85
  64. 64. 99% of all scores m = 100 -1 sd +1 sd +1s=130 +2 sd -1.96 s =70 -2.58 sd -1.96 sd -3s=55 +3 sd Here are the actual values -2.58 s =55 +1 s =115 -1 s =85
  65. 65. 99% of all scores m = 100 -1 sd +1 sd +1.96 s =130 -1.96 s =70 -2.58 sd -1.96 sd -3s=55 +3 sd +1.96 sd Here are the actual values -2.58 s =55 +1 s =115 -1 s =85
  66. 66. 99% of all scores m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 -1.96 s =70 -2.58 sd -1.96 sd +2.58 s =145 +2.58 sd +1.96 sd Here are the actual values -2.58 s =55
  67. 67. 99% of all scores m = 100 -2.58 sd -1.96 sd -1 sd +1 sd +1.96 sd +2.58 sd Based on the percentages of a normal distribution, we can insert the percentage of scores below each standard deviation point. +1 s =115 -1 s =85 +1.96 s =130 +2.58 s =145 -1.96 s =70 -2.58 s =55
  68. 68. 99% of all scores -2.58 sd -1.96 sd -1 sd +1 sd +1.96 sd +2.58 sd Attribute #4: percentages can be calculated below or above each standard deviation point in the distribution. m = 100 +1 s =115 -1 s =85 +1.96 s =130 +2.58 s =145 -1.96 s =70 -2.58 s =55
  69. 69. 99% of all scores m = 100 +1 s =115 -1 s =85 +1.96 s =130 +2.58 s =145 -1.96 s =70 -2.58 s =55 -2.58 sd -1.96 sd -1 sd +1 sd +1.96 sd +2.58 sd
  70. 70. 99% of all scores 95% of all scores m = 100 +1 s =115 -1 s =85 +1.96 s =130 +2.58 s =145 -1.96 s =70 -2.58 s =55 -2.58 sd -1.96 sd -1 sd +1 sd +1.96 sd +2.58 sd
  71. 71. 99% of all scores 95% of all scores 68% of all scores m = 100 +1 s =115 -1 s =85 +1.96 s =130 +2.58 s =145 -1.96 s =70 -2.58 s =55 -2.58 sd -1.96 sd -1 sd +1 sd +1.96 sd +2.58 sd
  72. 72. 99% of all scores 95% of all scores 68% of all scores m = 100 +1 s =115 -1 s =85 +1.96 s =130 -2.58 sd -1.96 sd -1 sd +1 sd +1.96 sd +2.58 sd With this information we can determine the probability that scores will fall into a number portions of the distribution. +2.58 s =145 -1.96 s =70 -2.58 s =55
  73. 73. For example:
  74. 74. There is a 0.5% chance that if you randomly selected a person that their IQ score would be below a 55 or a -2.58 SD 0.5% m = 100 +1 s =115 -1 s =85 +1.96 s =130 +2.58 s =145 -1.96 s =70 -2.58 s =55 -2.58 sd -1.96 sd -1 sd +1 sd +1.96 sd +2.58 sd
  75. 75. There is a 2.5% chance that if you randomly selected a person that their IQ score would be below a 70 or a -1.96 SD 2.5% m = 100 +1 s =115 -1 s =85 +1.96 s =130 +2.58 s =145 -1.96 s =70 -2.58 s =55 -2.58 sd -1.96 sd -1 sd +1 sd +1.96 sd +2.58 sd
  76. 76. There is a 16.5% chance that if you randomly selected a person that their IQ score would be below a 85 or a -1 SD 16.5% m = 100 +1 s =115 -1 s =85 +1.96 s =130 +2.58 s =145 -1.96 s =70 -2.58 s =55 -2.58 sd -1.96 sd -1 sd +1 sd +1.96 sd +2.58 sd
  77. 77. 50% There is a 50% chance that if you randomly selected a person that their IQ score would be below a 100 or a 0 SD m = 100 +1 s =115 -1 s =85 +1.96 s =130 +2.58 s =145 -1.96 s =70 -2.58 s =55 -2.58 sd -1.96 sd -1 sd +1 sd +1.96 sd +2.58 sd
  78. 78. There is a 50% chance that if you randomly selected a person that their IQ score would be above a 100 or a 0 SD 50% m = 100 +1 s =115 -1 s =85 +1.96 s =130 +2.58 s =145 -1.96 s =70 -2.58 s =55 -2.58 sd -1.96 sd -1 sd +1 sd +1.96 sd +2.58 sd
  79. 79. There is a 16.5% chance that if you randomly selected a person that their IQ score would be above a 115 or a +1 SD 16.5% m = 100 +1 s =115 -1 s =85 +1.96 s =130 +2.58 s =145 -1.96 s =70 -2.58 s =55 -2.58 sd -1.96 sd -1 sd +1 sd +1.96 sd +2.58 sd
  80. 80. m = 100 +1 s =115 -1 s =85 -1 sd +1 sd There is a 2.5% chance that if you randomly selected a person that their IQ score would be above a 130 or a +2.58 SD +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd 2.5% +2.58 s =145 +2.58 sd
  81. 81. m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd There is a 0.5% chance that if you randomly selected a person that their IQ score would be above a 145 or a +2.58 SD .5% +2.58 s =145 +2.58 sd
  82. 82. What you have just seen illustrated is the concept of probability density or the probability that a score or observation would be selected above, m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd below or between two points on a distribution.
  83. 83. Back to our example again.
  84. 84. m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd
  85. 85. m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd Here is the sample we randomly selected 풙 = 70
  86. 86. m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd The sample mean is 30 units away from the population mean (100 – 70 = 30). 풙 = 70
  87. 87. m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd The sample mean is 30 units away from the population mean (100 – 70 = 30). 풙 = 70 30
  88. 88. m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd Is that far away enough to be called statistically significantly different than the population? 풙 = 70
  89. 89. m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd How far is too far away? 풙 = 70
  90. 90. m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd Fortunately, statisticians have come up with a couple of distances that are considered too far away to be a part of the population. 풙 = 70
  91. 91. m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd These distances are measured in z-scores 풙 = 70
  92. 92. m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd 풙 = 70 Which are what these are: z scores
  93. 93. Let’s say statisticians determined that if the sample mean you collected is below a -1.96 z-score or above a +1.96 z-score that that’s just too far away from the mean to be a part of the population. m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd 풙 = 70
  94. 94. We know from previous slides that only 2.5% of the scores are below a -1.96 m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd 풙 = 70
  95. 95. We know from previous slides that only 2.5% of the scores are below a -1.96 m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd 풙 = 70 2.5%
  96. 96. Since anything at this point or below is considered to be too rare to be a part of this population, we would conclude that the population and the sample are statistically significantly different from one another. m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd 풙 = 70 2.5%
  97. 97. m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd 풙 = 70 And that’s our answer! 2.5%
  98. 98. What if the sample mean had been 105?
  99. 99. Since our decision rule is to determine that the sample mean is statistically significantly different than the population mean if the sample mean lies outside of the top or bottom 2.5% of all scores,
  100. 100. Since our decision rule is to determine that the sample mean is statistically significantly different than the population mean if the sample mean lies outside of the top or bottom 2.5% of all scores, m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd 2.5% 2.5%
  101. 101. Since our decision rule is to determine that the sample mean is statistically significantly different than the population mean if the sample mean lies outside of the top or bottom 2.5% of all scores, m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd 풙 = 105 2.5% 2.5%
  102. 102. . . . and the sample mean (105) does not lie in these outer regions, m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd 풙 = 105 2.5% 2.5%
  103. 103. Therefore, we would say that this is not a rare event and the probability that the sample is significantly similar to the population is high. m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd 풙 = 105 2.5% 2.5%
  104. 104. By the way, how do we figure out the z-score for an IQ score of 105.
  105. 105. We use the following formula to compute z-scores across the normal distribution:
  106. 106. We use the following formula to compute z-scores across the normal distribution: 풙 - m SD
  107. 107. We use the following formula to compute z-scores across the normal distribution: 풙 - m Here’s our SD sample mean: 70
  108. 108. We use the following formula to compute z-scores across the normal distribution: 70 - m Here’s our SD sample mean: 70
  109. 109. We use the following formula to compute z-scores across the normal distribution: 70 - m SD Here’s our Population mean: 100
  110. 110. We use the following formula to compute z-scores across the normal distribution: 70 - 100 SD Here’s our Population mean: 100
  111. 111. We use the following formula to compute z-scores across the normal distribution: 70 - 100 SD Here’s our Standard Deviation: 15
  112. 112. We use the following formula to compute z-scores across the normal distribution: 70 - 100 15 Here’s our Standard Deviation: 15
  113. 113. We use the following formula to compute z-scores across the normal distribution: 70 - 100 15
  114. 114. We use the following formula to compute z-scores across the normal distribution: 30 15
  115. 115. We use the following formula to compute z-scores across the normal distribution: 2.0
  116. 116. A z score of 2 is located right here m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd 풙 = 70 2.5% 2.5%
  117. 117. In some instances we may not know the population standard deviation s (in this case 15).
  118. 118. Without the standard deviation of the population we cannot determine the z-scores or the probability that a sample mean is too far away to be apart of the population.
  119. 119. Without the standard deviation of the population we cannot determine the z-scores or the probability that a sample mean is too far away to be apart of the population. m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd
  120. 120. Without the standard deviation of the population we cannot determine the z-scores or the probability that a sample mean is too far away to be apart of the population. m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd 50%
  121. 121. Without the standard deviation of the population we cannot determine the z-scores or the probability that a sample mean is too far away to be apart of the population. m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd 16.5%
  122. 122. Without the standard deviation of the population we cannot determine the z-scores or the probability that a sample mean is too far away to be apart of the population. m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd 2.5% +2.58 s =145 +2.58 sd
  123. 123. Without the standard deviation of the population we cannot determine the z-scores or the probability that a sample mean is too far away to be apart of the population. m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd Etc.
  124. 124. Therefore these values below cannot be computed: m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd
  125. 125. Therefore these values below cannot be computed: m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd
  126. 126. When we only know the population mean we use the Single Sample t-test.
  127. 127. Actually whenever we are dealing with a population and a sample, we generally use a single-sample t-test.
  128. 128. In the last example we relied on the population mean and standard deviation to determine if the sample mean was too far away from the population mean to be considered a part of the population.
  129. 129. The single sample t-test relies on a concept called the estimated standard error
  130. 130. The single sample t-test relies on a concept called the estimated standard error to compute something like a z-score to determine the probability distance between the population and the sample means.
  131. 131. We call it estimated because as you will see it is not really feasible to compute.
  132. 132. Standard error draws on two concepts:
  133. 133. 1. sampling distributions
  134. 134. 1. sampling distributions 2. t-distributions
  135. 135. Let’s begin with sampling distributions.
  136. 136. Let’s begin with sampling distributions. What you are about to see is purely theoretical, but it provides the justification for the formula we will use to run a single sample t-test.
  137. 137. Let’s begin with sampling distributions. What you are about to see is purely theoretical, but it provides the justification for the formula we will use to run a single sample t-test. x̄ – μ SEmean
  138. 138. x̄ – μ SEmean the mean of a sample
  139. 139. x̄ – μ SEmean the mean of a sample the mean of a population
  140. 140. x̄ – μ SEmean the mean of a sample the mean of a population the estimated standard error
  141. 141. x̄ – μ SEmean In our example, this is 70
  142. 142. 70̄ – μ SEmean In our example, this is 70
  143. 143. 70̄ – μ SEmean And this is 100
  144. 144. 70̄ – 100 SEmean And this is 100
  145. 145. The numerator here is easy to compute 70̄ – 100 SEmean
  146. 146. The numerator here is easy to compute -30 SEmean
  147. 147. -30 SEmean This value will help us know the distance between 70 and 100 in t-values
  148. 148. -30 SEmean If the estimated standard error is large, like 30, then the t value would be: -30/30 = -1
  149. 149. A -1.0 t-value is like a z-score as shown below:
  150. 150. A -1.0 t-value is like a z-score as shown below: m = 100 +1 s =115 -1 s =85 -1 sd +1 sd +1.96 s =130 +1.96 sd -1.96 s =70 -1.96 sd -2.58 s =55 -2.58 sd +2.58 s =145 +2.58 sd
  151. 151. But since we don’t know the population standard deviation, we have to use the standard deviation of the sample (not the population as we did before) to determine the distance in standard error units (or t values).
  152. 152. The focus of our theoretical justification is to explain our rationale for using information from the sample to compute the standard error or standard error of the mean.
  153. 153. The focus of our theoretical justification is to explain our rationale for using information from the sample to compute the standard error or standard error of the mean. x̄ – μ SEmean
  154. 154. Here comes the theory behind standard error:
  155. 155. Here comes the theory behind standard error: Imagine we took a sample of 20 IQ scores from the population.
  156. 156. Here comes the theory behind standard error: Imagine we took a sample of 20 IQ scores from the population. This sample of 20 would have its own IQ mean, standard deviation and distribution.
  157. 157. Here comes the theory behind standard error: Imagine we took a sample of 20 IQ scores from the population. This sample of 20 would have its own IQ mean, standard deviation and distribution. x ̄= 70 SD = 10
  158. 158. Here comes the theory behind standard error: Imagine we took a sample of 20 IQ scores from the population. This sample of 20 would have its own IQ mean, standard deviation and distribution. And then let’s say we took another sample of 20 with its mean and distribution, x ̄= 70
  159. 159. Here comes the theory behind standard error: Imagine we took a sample of 20 IQ scores from the population. This sample of 20 would have its own IQ mean, standard deviation and distribution. And then let’s say we took another sample of 20 with its mean and distribution, x ̄= 70 x ̄= 100
  160. 160. Here comes the theory behind standard error: Imagine we took a sample of 20 IQ scores from the population. This sample of 20 would have its own IQ mean, standard deviation and distribution. And then let’s say we took another sample of 20 with its mean and distribution, and another x ̄= 70 x ̄= 100
  161. 161. Here comes the theory behind standard error: Imagine we took a sample of 20 IQ scores from the population. This sample of 20 would have its own IQ mean, standard deviation and distribution. And then let’s say we took another sample of 20 with its mean and distribution, and another x ̄= 70 x ̄= 100 x ̄= 120
  162. 162. Here comes the theory behind standard error: Imagine we took a sample of 20 IQ scores from the population. This sample of 20 would have its own IQ mean, standard deviation and distribution. And then let’s say we took another sample of 20 with its mean and distribution, and another, and another x ̄= 70 x ̄= 100 x ̄= 120 x ̄= 140
  163. 163. Here comes the theory behind standard error: Imagine we took a sample of 20 IQ scores from the population. This sample of 20 would have its own IQ mean, standard deviation and distribution. And then let’s say we took another sample of 20 with its mean and distribution, and another, and another, and so on… x ̄= 70 x ̄= 100 x ̄= 120 x ̄= 140
  164. 164. Let’s say, theoretically, that we do this one hundred times.
  165. 165. Let’s say, theoretically, that we do this one hundred times. We now have 100 samples of 20 person IQ scores:
  166. 166. Let’s say, theoretically, that we do this one hundred times. We now have 100 samples of 20 person IQ scores:
  167. 167. Let’s say, theoretically, that we do this one hundred times. We now have 100 samples of 20 person IQ scores: We take the mean of each of those samples: x ̄= 115 x ̄= 100 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 110 x ̄= 102 x ̄= 120 x ̄= 90 x ̄= 114 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 95
  168. 168. And we create a new distribution called the sampling distribution of the means x ̄= 115 x ̄= 100 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 110 x ̄= 102 x ̄= 120 x ̄= 90 x ̄= 114 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 95
  169. 169. And x ̄= 115 x ̄= 105 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90 x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95
  170. 170. And x ̄= 115 x ̄= 105 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90 x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95 70 75 70 85 90 95 100 105 110 115 120 125
  171. 171. And x ̄= 115 x ̄= 105 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90 x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95 70 75 70 85 90 95 100 105 110 115 120 125
  172. 172. And x ̄= 115 x ̄= 105 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90 x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95 70 75 70 85 90 95 100 105 110 115 120 125
  173. 173. And x ̄= 115 x ̄= 105 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90 x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95 70 75 70 85 90 95 100 105 110 115 120 125
  174. 174. And x ̄= 115 x ̄= 105 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90 x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95 70 75 70 85 90 95 100 105 110 115 120 125
  175. 175. And x ̄= 115 x ̄= 105 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90 x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95 70 75 70 85 90 95 100 105 110 115 120 125
  176. 176. And x ̄= 115 x ̄= 105 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90 x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95 70 75 70 85 90 95 100 105 110 115 120 125
  177. 177. And x ̄= 115 x ̄= 105 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90 x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95 70 75 70 85 90 95 100 105 110 115 120 125
  178. 178. And x ̄= 115 x ̄= 105 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90 x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95 70 75 70 85 90 95 100 105 110 115 120 125
  179. 179. And x ̄= 115 x ̄= 105 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90 x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95 70 75 70 85 90 95 100 105 110 115 120 125
  180. 180. And x ̄= 115 x ̄= 105 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90 x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95 70 75 70 85 90 95 100 105 110 115 120 125
  181. 181. And x ̄= 115 x ̄= 105 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90 x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95 70 75 70 85 90 95 100 105 110 115 120 125
  182. 182. And x ̄= 115 x ̄= 105 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90 x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95 70 75 70 85 90 95 100 105 110 115 120 125
  183. 183. And x ̄= 115 x ̄= 105 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90 x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95 70 75 70 85 90 95 100 105 110 115 120 125
  184. 184. And x ̄= 115 x ̄= 105 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90 x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95 70 75 70 85 90 95 100 105 110 115 120 125
  185. 185. And x ̄= 115 x ̄= 105 x ̄= 100 x ̄= 120 x ̄= 90 etc. … x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90 x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95 70 75 70 85 90 95 100 105 110 115 120 125
  186. 186. And then we do something interesting. We take the standard deviation of this sampling distribution.
  187. 187. And then we do something interesting. We take the standard deviation of this sampling distribution. If these sample means are close to one another then the standard deviation will be small.
  188. 188. And then we do something interesting. We take the standard deviation of this sampling distribution. If these sample means are close to one another then the standard deviation will be small. 70 75 70 85 90 95 100 105 110 115 120 125
  189. 189. And then we do something interesting. We take the standard deviation of this sampling distribution. If these sample means are far apart from one another then the standard deviation will be large. 70 75 70 85 90 95 100 105 110 115 120 125
  190. 190. This standard deviation of the sampling distribution of the means has another name:
  191. 191. This standard deviation of the sampling distribution of the means has another name: the standard error.
  192. 192. This standard deviation of the sampling distribution of the means has another name: the standard error. x̄ – μ SEmean the estimated standard error
  193. 193. Standard error is a unit of measurement that makes it possible to determine if a raw score difference is really significant or not.
  194. 194. Standard error is a unit of measurement that makes it possible to determine if a raw score difference is really significant or not. Think of it this way. If you get a 92 on a 100 point test and the general population gets on average a 90, is there really a significant difference between you and the population at large? If you retook the test over and over again would you likely outperform or underperform their average of 90?
  195. 195. Standard error is a unit of measurement that makes it possible to determine if a raw score difference is really significant or not. Think of it this way. If you get a 92 on a 100 point test and the general population gets on average a 90, is there really a significant difference between you and the population at large? If you retook the test over and over again would you likely outperform or underperform their average of 90? Standard error helps us understand the likelihood that those results would replicate the same way over and over again . . . or not.
  196. 196. So let’s say we calculate the standard error to be 0.2. Using the formula below we will determine how many standard error units you are apart from each other.
  197. 197. So let’s say we calculate the standard error to be 0.2. Using the formula below we will determine how many standard error units you are apart from each other. x̄ – μ SEmean t =
  198. 198. So let’s say we calculate the standard error to be 0.2. Using the formula below we will determine how many standard error units you are apart from each other. 92 – 90 0.2 your score t = standard error population average
  199. 199. So let’s say we calculate the standard error to be 0.2. Using the formula below we will determine how many standard error units you are apart from each other. 2 0.2 t =
  200. 200. So let’s say we calculate the standard error to be 0.2. Using the formula below we will determine how many standard error units you are apart from each other. 2 0.2 t = This is the raw score difference
  201. 201. So let’s say we calculate the standard error to be 0.2. Using the formula below we will determine how many standard error units you are apart from each other. t = 10.0 And this is the difference in standard error units
  202. 202. So let’s say we calculate the standard error to be 0.2. Using the formula below we will determine how many standard error units you are apart from each other. t = 10.0 And this is the difference in standard error units, otherwise known as a t statistic or t value
  203. 203. So, while your test score is two raw scores above the average for the population, you are 10.0 standard error units higher than the population score.
  204. 204. So, while your test score is two raw scores above the average for the population, you are 10.0 standard error units higher than the population score. While 2.0 raw scores do not seem like a lot, 10.0 standard error units constitute a big difference!
  205. 205. So, while your test score is two raw scores above the average for the population, you are 10.0 standard error units higher than the population score. While 2.0 raw scores do not seem like a lot, 10.0 standard error units constitute a big difference! This means that this result is most likely to replicate and did not happen by chance.
  206. 206. But what if the standard error were much bigger, say, 4.0?
  207. 207. But what if the standard error were much bigger, say, 4.0? x̄ – μ SEmean t =
  208. 208. But what if the standard error were much bigger, say, 4.0? 92 – 90 4.0 your score t = standard error population average
  209. 209. But what if the standard error were much bigger, say, 4.0? 2 4.0 t =
  210. 210. But what if the standard error were much bigger, say, 4.0? t = 0.5
  211. 211. Once again, your raw score difference is still 2.0 but you are only 0.5 standard error units apart. That distance is most likely too small to be statistically significantly different if replicated over a hundred times.
  212. 212. Once again, your raw score difference is still 2.0 but you are only 0.5 standard error units apart. That distance is most likely too small to be statistically significantly different if replicated over a hundred times. We will show you how to determine when the number of standard error units is significantly different or not.
  213. 213. One more example:
  214. 214. One more example: Let’s say the population average on the test is 80 and you still received a 92. But the standard error is 36.0.
  215. 215. One more example: Let’s say the population average on the test is 80 and you still received a 92. But the standard error is 36.0. Let’s do the math again.
  216. 216. One more example: Let’s say the population average on the test is 80 and you still received a 92. But the standard error is 36.0. Let’s do the math again. x̄ – μ SEmean t =
  217. 217. One more example: Let’s say the population average on the test is 80 and you still received a 92. But the standard error is 36.0. Let’s do the math again. 92 – 80 36.0 your score t = standard error population average
  218. 218. One more example: Let’s say the population average on the test is 70 and you still received a 92. But the standard error is 36.0. Let’s do the math again. 12 36.0 t =
  219. 219. One more example: Let’s say the population average on the test is 70 and you still received a 92. But the standard error is 36.0. Let’s do the math again. t = 0.3
  220. 220. In this case, while your raw score is 12 points higher (a large amount), you are only 0.3 standard error units higher.
  221. 221. In this case, while your raw score is 12 points higher (a large amount), you are only 0.3 standard error units higher. The standard error is so large (36.0) that if you were to take the test 1000 times with no growth in between it is most likely that your scores would vary greatly (92 on one day, 77 on another day, 87 on another day and so on and so forth.)
  222. 222. In summary, the single sample test t value is the number of standard error units that separate the sample mean from the population mean:
  223. 223. In summary, the single sample test t value is the number of standard error units that separate the sample mean from the population mean: x̄ – μ SEmean t = Let’s see this play out with our original example.
  224. 224. Let’s say our of 20 has an average IQ score of 70.
  225. 225. Let’s say our of 20 has an average IQ score of 70. x̄ – μ SEmean t =
  226. 226. Let’s say our of 20 has an average IQ score of 70. 70 – μ SEmean t =
  227. 227. We already know that the population mean is 100. 70 – μ SEmean t =
  228. 228. We already know that the population mean is 100. 70 – 100 SEmean t =
  229. 229. Let’s say the Standard Error of the Sampling Distribution means is 5 70 – 100 SEmean t =
  230. 230. Let’s say the Standard Error of the Sampling Distribution means is 5 70 – 100 5 t =
  231. 231. The t value would be: 70 – 100 5 t =
  232. 232. The t value would be: -30 5 t =
  233. 233. The t value would be: t = -6
  234. 234. So all of this begs the question: How do I know if a t value of -6 is rare or common?
  235. 235. So all of this begs the question: How do I know if a t value of -6 is rare or common? • If it is rare we can accept the null hypothesis and say that there is a difference between the sample and the population.
  236. 236. So all of this begs the question: How do I know if a t value of -6 is rare or common? • If it is rare we can accept the null hypothesis and say that there is a difference between the sample and the population. • If it is common we can reject the null hypothesis and say there is not a significant difference between veggie eating IQ scores and the general population IQ scores (which in this unique case is what we want)
  237. 237. So all of this begs the question: How do I know if a t value of 5 is rare or common? • If it is rare we can accept the null hypothesis and say that there is a difference between the sample and the population. • If it is common we can reject the null hypothesis and say there is not a significant difference between veggie eating IQ scores and the general population IQ scores (which in this unique case is what we want) Here is what we do: We compare this value (6) with the critical t value.
  238. 238. What is the critical t value?
  239. 239. What is the critical t value? The critical t value is a point in the distribution that arbitrarily separates the common from the rare occurrences.
  240. 240. What is the critical t value? The critical t value is a point in the distribution that arbitrarily separates the common from the rare occurrences.
  241. 241. What is the critical t value? The critical t value is a point in the distribution that arbitrarily separates the common from the rare occurrences. rare occurrence
  242. 242. What is the critical t value? The critical t value is a point in the distribution that arbitrarily separates the common from the rare occurrences. rare occurrence rare occurrence
  243. 243. What is the critical t value? The critical t value is a point in the distribution that arbitrarily separates the common from the rare occurrences. rare occurrence common occurrence rare occurrence
  244. 244. What is the critical t value? The critical t value is a point in the distribution that arbitrarily separates the common from the rare occurrences. rare occurrence common occurrence rare occurrence This represents the rare/common possibilities used to determine if the sample mean is similar the population mean).
  245. 245. If this were a normal distribution the red line would have a z critical value of + or – 1.96 rare occurrence common occurrence rare occurrence
  246. 246. If this were a normal distribution the red line would have a z critical value of + or – 1.96 rare occurrence common occurrence rare occurrence
  247. 247. If this were a normal distribution the red line would have a z critical value of + or – 1.96 rare occurrence common occurrence rare occurrence
  248. 248. If this were a normal distribution the red line would have a z critical value of + or – 1.96 (which is essentially a t critical value but for a normal distribution.) rare occurrence common occurrence rare occurrence
  249. 249. A t or z value of + or – 1.96 means that 95% of the scores are in the center of the distribution and 5% are to the left and right of it.
  250. 250. A t or z value of + or – 1.96 means that 95% of the scores are in the center of the distribution and 5% are to the left and right of it. common occurrence rare occurrence rare occurrence
  251. 251. A t or z value of + or – 1.96 means that 95% of the scores are in the center of the distribution and 5% are to the left and right of it. rare occurrence common occurrence rare occurrence -1.96 +1.96
  252. 252. A t or z value of + or – 1.96 means that 95% of the scores are in the center of the distribution and 5% are to the left and right of it. rare occurrence common occurrence rare occurrence -1.96 +1.96
  253. 253. A t or z value of + or – 1.96 means that 95% of the scores are in the center of the distribution and 5% are to the left and right of it. rare occurrence common occurrence rare occurrence -1.96 +1.96
  254. 254. A t or z value of + or – 1.96 means that 95% of the scores are in the center of the distribution and 5% are to the left and right of it. rare occurrence common occurrence rare occurrence -1.96 +1.96
  255. 255. A t or z value of + or – 1.96 means that 95% of the scores are in the center of the distribution and 5% are to the left and right of it. rare occurrence common occurrence rare occurrence -1.96 +1.96
  256. 256. So if the t value computed from this equation:
  257. 257. So if the t value computed from this equation: x̄ – μ SEmean t =
  258. 258. So if the t value computed from this equation: x̄ – μ SEmean t = … is between -1.96 and +1.96 we would say that that result is not rare and we would fail to reject the null hypothesis.
  259. 259. So if the t value computed from this equation: x̄ – μ SEmean t = … is between -1.96 and +1.96 we would say that that result is not rare and we would fail to reject the null hypothesis. rare occurrence common occurrence rare occurrence -1.96 +1.96
  260. 260. However, if the t value is smaller than -1.96 or larger than +1.96 we would say that that result is rare and we would reject the null hypothesis.
  261. 261. However, if the t value is smaller than -1.96 or larger than +1.96 we would say that that result is rare and we would reject the null hypothesis. rare occurrence common occurrence rare occurrence -1.96 +1.96
  262. 262. So if the t value computed from this equation:
  263. 263. So if the t value computed from this equation: x̄ – μ SEmean t =
  264. 264. So if the t value computed from this equation: x̄ – μ SEmean t = … is between -1.96 and +1.96 we would say that that result is not rare and we would fail to reject the null hypothesis.
  265. 265. As a review, let’s say the distribution is normal. When the distribution is normal and we want to locate the z critical for a two tailed test at a level of significance of .05 (which means that if we took 100 samples we are willing to be wrong 5 times and still reject the null hypothesis), the z critical would be -+1.96.
  266. 266. As a review, let’s say the distribution is normal. When the distribution is normal and we want to locate the z critical for a two tailed test at a level of significance of .05 (which means that if we took 100 samples we are willing to be wrong 5 times and still reject the null hypothesis), the z critical would be -+1.96. - 1.96 + 1.96
  267. 267. Because we generally do not have the resources to take 100 IQ samples of those who eat veggies, we have to estimate the t value from one sample.
  268. 268. Because we generally do not have the resources to take 100 IQ samples of those who eat veggies, we have to estimate the t value from one sample. x ̄= 115 x ̄= 105 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90 x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95
  269. 269. Because we generally do not have the resources to take 100 IQ samples of those who eat veggies, we have to estimate the t value from one sample. x ̄= 115 x ̄= 105 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 110 x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90 x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95
  270. 270. Because we generally do not have the resources to take 100 IQ samples of those who eat veggies, we have to estimate the t value from one sample. x ̄= 110
  271. 271. Because we generally do not have the resources to take 100 IQ samples of those who eat veggies, we have to estimate the t value from one sample. So let’s imagine we selected a sample of 20 veggie eaters with an average IQ score of 70. x ̄= 70
  272. 272. Because we generally do not have the resources to take 100 IQ samples of those who eat veggies, we have to estimate the t value from one sample. So let’s imagine we selected a sample x ̄= 70 of 20 veggie eaters with an average IQ score of 70. Because we did not take hundreds of samples of 20 veggie eaters each, average each sample’s IQ scores, and form a sampling distribution from which we could compute the standard error and then the t value, we have to figure out another way to compute an estimate of the standard error.
  273. 273. Because we generally do not have the resources to take 100 IQ samples of those who eat veggies, we have to estimate the t value from one sample. So let’s imagine we selected a sample x ̄= 70 of 20 veggie eaters with an average IQ score of 70. Because we did not take hundreds of samples of 20 veggie eaters each, average each sample’s IQ scores, and form a sampling distribution from which we could compute the standard error and then the t value, we have to figure out another way to compute an estimate of the standard error. There is another way.
  274. 274. Since it is not practical to collect a hundreds of samples of 20 from the population, compute their mean score and calculate the standard error, we must estimate it using the following equation: S n SEmean =
  275. 275. Since it is not practical to collect a hundreds of samples of 20 from the population, compute their mean score and calculate the standard error, we must estimate it using the following equation: S n SEmean = Standard Deviation of the sample
  276. 276. We estimate the standard error using the following equation: S n SEmean = Standard Deviation of the sample
  277. 277. We estimate the standard error using the following equation: SEmean = Standard Deviation of the sample Square root of the sample size S n We won’t go into the derivation of this formula, but just know that this acts as a good substitute in the place of taking hundreds of samples and computing the standard deviation to get the actual standard error.
  278. 278. So remember this point: The equation below is an estimate of the standard error, not the actual standard error, because the actual standard error is not feasible to compute.
  279. 279. So remember this point: The equation below is an estimate of the standard error, not the actual standard error, because the actual standard error is not feasible to compute. S n SEmean =
  280. 280. So remember this point: The equation below is an estimate of the standard error, not the actual standard error, because the actual standard error is not feasible to compute. S n SEmean = However, just know that when researchers have taken hundreds of samples and computed their means and then taken the standard deviation of all of those means they come out pretty close to one another.
  281. 281. Now comes another critical point dealing with t-distributions.
  282. 282. Now comes another critical point dealing with t-distributions. Because we are working with a sample that is generally small, we do not use the same normal distribution to determine the critical z or t (-+1.96).
  283. 283. Now comes another critical point dealing with t-distributions. Because we are working with a sample that is generally small, we do not use the same normal distribution to determine the critical z or t (-+1.96). Here is what the z distribution looks like for samples generally larger than 30:
  284. 284. Now comes another critical point dealing with t-distributions. Because we are working with a sample that is generally small, we do not use the same normal distribution to determine the critical z or t (-+1.96). Here is what the z distribution looks like for samples generally larger than 30: 95% of the scores - 1.96 + 1.96 Sample Size 30+
  285. 285. As the sample size decreases the critical values increase - making it harder to get significance.
  286. 286. As the sample size decreases the critical values increase - making it harder to get significance. Notice how the t distribution gets shorter and wider when the sample size is smaller. Notice also how the t critical values increase as well. 95% of the scores Sample Size 20 - 2.09 + 2.09
  287. 287. As the sample size decreases the critical values increase - making it harder to get significance. Notice how the t distribution gets shorter and wider when the sample size is smaller. Notice also how the t critical values increase as well. 95% of the scores Sample Size 10 - 2.26 + 2.26
  288. 288. As the sample size decreases the critical values increase - making it harder to get significance. Notice how the t distribution gets shorter and wider when the sample size is smaller. Notice also how the t critical values increase as well. 95% of the scores Sample Size 5 - 2.78 + 2.78
  289. 289. To determine the critical t we need two values:
  290. 290. To determine the critical t we need two values: • the degrees of freedom (sample size minus one [20-1 = 19])
  291. 291. To determine the critical t we need two values: • the degrees of freedom (sample size minus one [20-1 = 19]) • the significance level (.05 or .025 for two tailed test)
  292. 292. To determine the critical t we need two values: • the degrees of freedom (sample size minus one [20-1 = 19]) • the significance level (.05 or .025 for two tailed test) Using these two values we can locate the critical t
  293. 293. So, our t critical value that separates the common from the rare occurrences in this case is + or – 2.09
  294. 294. So, our t critical value that separates the common from the rare occurrences in this case is + or – 2.09 95% of the scores Sample Size 20 - 2.09 + 2.09
  295. 295. What was our calculated t value again?
  296. 296. What was our calculated t value again? x̄ – μ SEmean t =
  297. 297. What was our calculated t value again? 70 – μ SEmean t =
  298. 298. We already know that the population mean is 100. 70 – μ SEmean t =
  299. 299. We already know that the population mean is 100. 70 – 100 SEmean t =
  300. 300. We already know that the population mean is 100. 70 – 100 SEmean t = To calculate the estimated standard error of the mean distribution we use the following equation:
  301. 301. We already know that the population mean is 100. 70 – 100 SEmean t = To calculate the estimated standard error of the mean distribution we use the following equation: S n SEmean =
  302. 302. We already know that the population mean is 100. 70 – 100 SEmean t = The Standard Deviation for this sample is 26.82 and the sample size, of course, is 20 S n SEmean =
  303. 303. We already know that the population mean is 100. 70 – 100 SEmean t = Let’s plug in those values: S n SEmean =
  304. 304. We already know that the population mean is 100. 70 – 100 SEmean t = Let’s plug in those values: 26.82 20 SEmean =
  305. 305. We already know that the population mean is 100. 70 – 100 SEmean 26.82 4.47 t = Let’s plug in those values: SEmean =
  306. 306. We already know that the population mean is 100. 70 – 100 SEmean 6.0 t = Let’s plug in those values: SEmean =
  307. 307. We already know that the population mean is 100. 70 – 100 6.0 6.0 t = Let’s plug in those values: SEmean =
  308. 308. We already know that the population mean is 100. 70 – 100 6.0 t =
  309. 309. Now do the math: 70 – 100 6.0 t =
  310. 310. The t value would be: 70 – 100 6.0 t =
  311. 311. The t value would be: -30 6.0 t =
  312. 312. The t value would be: t = -5
  313. 313. With a t value of -5 we are ready to compare it to the critical t:
  314. 314. With a t value of -5 we are ready to compare it to the critical t:
  315. 315. With a t value of -5 we are ready to compare it to the critical t: 2.093
  316. 316. With a t value of -5 we are ready to compare it to the critical t: 2.093 or 2.093 below the mean or -2.093.
  317. 317. Since a t value of +5.0 is below the cutoff point (critical t) of -2.09, we would reject the null hypothesis
  318. 318. Since a t value of +5.0 is below the cutoff point (critical t) of -2.09, we would reject the null hypothesis 95% of the scores - 2.09 + 2.09
  319. 319. Here is how we would state our results:
  320. 320. Here is how we would state our results: The randomly selected sample of twenty IQ scores with a sample mean of 70 is statistically significantly different then the population of IQ scores.
  321. 321. Now, what if the standard error had been much larger, say, 15.0 instead of 6.0?
  322. 322. Now, what if the standard error had been much larger, say, 15.0 instead of 6.0? 70 – 100 6.0 t =
  323. 323. Now, what if the standard error had been much larger, say, 15.0 instead of 6.0? 70 – 100 6.0 t = 70 – 100 15.0 t =
  324. 324. Now, what if the standard error had been much larger, say, 15.0 instead of 6.0? 70 – 100 6.0 t = 30 15.0 t =
  325. 325. Now, what if the standard error had been much larger, say, 10.0 instead of 2.0? 70 – 100 t = t = 2.0 6.0
  326. 326. A t value of -2.0 is not below the cutoff point (critical t) of -2.09 and would be considered a common rather than a rare outcome. We therefore would fail to reject the null hypothesis
  327. 327. A t value of -2.0 is not below the cutoff point (critical t) of -2.09 and would be considered a common rather than a rare outcome. We therefore would fail to reject the null hypothesis 95% of the scores - 2.09 - 2.0 + 2.09
  328. 328. So in summary, the single sample t-test helps us determine the probability that the difference between a sample and a population did or did not occur by chance.
  329. 329. So in summary, the single sample t-test helps us determine the probability that the difference between a sample and a population did or did not occur by chance. It utilizes the concepts of standard error, common and rare occurrences, and t-distributions to justify its use.
  330. 330. End of Presentation

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