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Unit 4: Inference for numerical variables Lecture 2: t -distribution Statistics 101 Thomas Leininger June 5, 2013 Small sample inference for the mean Small sample inference for the mean 1 Introducing the t distribution Evaluating hypotheses

  1. Small sample inference for the mean Introducing the t distribution Small sample inference for the mean 1 Introducing the t distribution Evaluating hypotheses using the t distribution Constructing confidence intervals using the t distribution Synthesis The t distribution for the difference of two means 2 Sampling distribution for the difference of two means Hypothesis testing for the difference of two means Confidence intervals for the difference of two means Recap Statistics 101 U4 - L2: t -distribution Thomas Leininger

  2. Small sample inference for the mean Introducing the t distribution The t distribution When working with small samples and with σ unknown (almost always), the uncertainty of the standard error estimate is addressed by using a new distribution: the t distribution . Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 7 / 33

  3. Small sample inference for the mean Introducing the t distribution The t distribution When working with small samples and with σ unknown (almost always), the uncertainty of the standard error estimate is addressed by using a new distribution: the t distribution . This distribution also has a bell shape, but its tails are thicker than the normal model’s. Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 7 / 33

  4. Small sample inference for the mean Introducing the t distribution The t distribution When working with small samples and with σ unknown (almost always), the uncertainty of the standard error estimate is addressed by using a new distribution: the t distribution . This distribution also has a bell shape, but its tails are thicker than the normal model’s. Therefore observations are more likely to fall beyond two SDs from the mean than under the normal distribution. normal t −4 −2 0 2 4 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 7 / 33

  5. Small sample inference for the mean Introducing the t distribution The t distribution (cont.) Always centered at zero and symmetric, like the standard normal ( z ) distribution. Has a single parameter: degrees of freedom ( df ). normal t, df=10 t, df=5 t, df=2 t, df=1 −2 0 2 4 6 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 8 / 33

  6. Small sample inference for the mean Introducing the t distribution The t distribution (cont.) Always centered at zero and symmetric, like the standard normal ( z ) distribution. Has a single parameter: degrees of freedom ( df ). normal t, df=10 t, df=5 t, df=2 t, df=1 −2 0 2 4 6 What happens to shape of the t distribution as df increases? Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 8 / 33

  7. Small sample inference for the mean Introducing the t distribution The t distribution (cont.) Always centered at zero and symmetric, like the standard normal ( z ) distribution. Has a single parameter: degrees of freedom ( df ). normal t, df=10 t, df=5 t, df=2 t, df=1 −2 0 2 4 6 What happens to shape of the t distribution as df increases? Approaches normal. Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 8 / 33

  8. Small sample inference for the mean Evaluating hypotheses using the t distribution Small sample inference for the mean 1 Introducing the t distribution Evaluating hypotheses using the t distribution Constructing confidence intervals using the t distribution Synthesis The t distribution for the difference of two means 2 Sampling distribution for the difference of two means Hypothesis testing for the difference of two means Confidence intervals for the difference of two means Recap Statistics 101 U4 - L2: t -distribution Thomas Leininger

  9. Small sample inference for the mean Evaluating hypotheses using the t distribution Back to Friday the 13 th 6 th 13 th type date diff location 1 traffic 1990, July 139246 138548 698 loc 1 2 traffic 1990, July 134012 132908 1104 loc 2 3 traffic 1991, September 137055 136018 1037 loc 1 4 traffic 1991, September 133732 131843 1889 loc 2 5 traffic 1991, December 123552 121641 1911 loc 1 6 traffic 1991, December 121139 118723 2416 loc 2 7 traffic 1992, March 128293 125532 2761 loc 1 8 traffic 1992, March 124631 120249 4382 loc 2 9 traffic 1992, November 124609 122770 1839 loc 1 10 traffic 1992, November 117584 117263 321 loc 2 ↓ x di ff = 1836 ¯ s di ff = 1176 n = 10 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 9 / 33

  10. Small sample inference for the mean Evaluating hypotheses using the t distribution Finding the test statistic Test statistic for inference on a small sample mean The test statistic for inference on a small sample ( n < 50 ) mean is the T statistic with df = n − 1 . T df = point estimate − null value SE Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 10 / 33

  11. Small sample inference for the mean Evaluating hypotheses using the t distribution Finding the test statistic Test statistic for inference on a small sample mean The test statistic for inference on a small sample ( n < 50 ) mean is the T statistic with df = n − 1 . T df = point estimate − null value SE in context... point estimate ¯ x di ff = 1836 = Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 10 / 33

  12. Small sample inference for the mean Evaluating hypotheses using the t distribution Finding the test statistic Test statistic for inference on a small sample mean The test statistic for inference on a small sample ( n < 50 ) mean is the T statistic with df = n − 1 . T df = point estimate − null value SE in context... point estimate x di ff = 1836 ¯ = s di ff √ n = 1176 SE = 372 √ = 10 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 10 / 33

  13. Small sample inference for the mean Evaluating hypotheses using the t distribution Finding the test statistic Test statistic for inference on a small sample mean The test statistic for inference on a small sample ( n < 50 ) mean is the T statistic with df = n − 1 . T df = point estimate − null value SE in context... point estimate ¯ x di ff = 1836 = s di ff √ n = 1176 SE = 372 √ = 10 1836 − 0 T = 4 . 94 = 372 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 10 / 33

  14. Small sample inference for the mean Evaluating hypotheses using the t distribution Finding the test statistic Test statistic for inference on a small sample mean The test statistic for inference on a small sample ( n < 50 ) mean is the T statistic with df = n − 1 . T df = point estimate − null value SE in context... point estimate ¯ x di ff = 1836 = s di ff √ n = 1176 SE = 372 √ = 10 1836 − 0 T = 4 . 94 = 372 df 10 − 1 = 9 = Note: Null value is 0 because in the null hypothesis we set µ di ff = 0 . Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 10 / 33

  15. Small sample inference for the mean Evaluating hypotheses using the t distribution Finding the p-value The p-value is, once again, calculated as the area tail area under the t distribution. Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 11 / 33

  16. Small sample inference for the mean Evaluating hypotheses using the t distribution Finding the p-value The p-value is, once again, calculated as the area tail area under the t distribution. Using R: > 2 * pt(4.94, df = 9, lower.tail = FALSE) [1] 0.0008022394 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 11 / 33

  17. Small sample inference for the mean Evaluating hypotheses using the t distribution Finding the p-value The p-value is, once again, calculated as the area tail area under the t distribution. Using R: > 2 * pt(4.94, df = 9, lower.tail = FALSE) [1] 0.0008022394 Using a web applet: http://www.socr.ucla.edu/htmls/SOCR Distributions.html Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 11 / 33

  18. Small sample inference for the mean Evaluating hypotheses using the t distribution Finding the p-value The p-value is, once again, calculated as the area tail area under the t distribution. Using R: > 2 * pt(4.94, df = 9, lower.tail = FALSE) [1] 0.0008022394 Using a web applet: http://www.socr.ucla.edu/htmls/SOCR Distributions.html Or when these aren’t available, we can use a t table. Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 11 / 33

  19. Small sample inference for the mean Evaluating hypotheses using the t distribution Finding the p-value Locate the calculated T statistic on the appropriate df row, obtain the p-value from the corresponding column heading (one or two tail, depending on the alternative hypothesis). one tail 0.100 0.050 0.025 0.010 0.005 two tails 0.200 0.100 0.050 0.020 0.010 1 3.08 6.31 12.71 31.82 63.66 df 2 1.89 2.92 4.30 6.96 9.92 3 1.64 2.35 3.18 4.54 5.84 . . . . . . . . . . . . . . . 17 1.33 1.74 2.11 2.57 2.90 18 1.33 1.73 2.10 2.55 2.88 19 1.33 1.73 2.09 2.54 2.86 20 1.33 1.72 2.09 2.53 2.85 . . . . . . . . . . . . . . . 400 1.28 1.65 1.97 2.34 2.59 500 1.28 1.65 1.96 2.33 2.59 ∞ 1.28 1.64 1.96 2.33 2.58 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 12 / 33

  20. Small sample inference for the mean Evaluating hypotheses using the t distribution Finding the p-value (cont.) one tail 0.100 0.050 0.025 0.010 0.005 two tails 0.200 0.100 0.050 0.020 0.010 df 6 1.44 1.94 2.45 3.14 3.71 7 1.41 1.89 2.36 3.00 3.50 8 1.40 1.86 2.31 2.90 3.36 9 1.38 1.83 2.26 2.82 3.25 10 1.37 1.81 2.23 2.76 3.17 T = 4 . 94 df = 9 µ diff = 0 x diff = 1836 − 1836 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 13 / 33

  21. Small sample inference for the mean Evaluating hypotheses using the t distribution Finding the p-value (cont.) one tail 0.100 0.050 0.025 0.010 0.005 two tails 0.200 0.100 0.050 0.020 0.010 df 6 1.44 1.94 2.45 3.14 3.71 7 1.41 1.89 2.36 3.00 3.50 8 1.40 1.86 2.31 2.90 3.36 9 1.38 1.83 2.26 2.82 3.25 10 1.37 1.81 2.23 2.76 3.17 T = 4 . 94 df = 9 µ diff = 0 x diff = 1836 − 1836 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 13 / 33

  22. Small sample inference for the mean Evaluating hypotheses using the t distribution Finding the p-value (cont.) one tail 0.100 0.050 0.025 0.010 0.005 two tails 0.200 0.100 0.050 0.020 0.010 df 6 1.44 1.94 2.45 3.14 3.71 7 1.41 1.89 2.36 3.00 3.50 8 1.40 1.86 2.31 2.90 3.36 9 1.38 1.83 2.26 2.82 3.25 10 1.37 1.81 2.23 2.76 3.17 T = 4 . 94 df = 9 µ diff = 0 x diff = 1836 − 1836 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 13 / 33

  23. Small sample inference for the mean Evaluating hypotheses using the t distribution Finding the p-value (cont.) one tail 0.100 0.050 0.025 0.010 0.005 two tails 0.200 0.100 0.050 0.020 0.010 df 6 1.44 1.94 2.45 3.14 3.71 7 1.41 1.89 2.36 3.00 3.50 8 1.40 1.86 2.31 2.90 3.36 9 1.38 1.83 2.26 2.82 3.25 10 1.37 1.81 2.23 2.76 3.17 T = 4 . 94 df = 9 What is the conclusion of the hy- pothesis test? µ diff = 0 x diff = 1836 − 1836 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 13 / 33

  24. Small sample inference for the mean Evaluating hypotheses using the t distribution Finding the p-value (cont.) one tail 0.100 0.050 0.025 0.010 0.005 two tails 0.200 0.100 0.050 0.020 0.010 df 6 1.44 1.94 2.45 3.14 3.71 7 1.41 1.89 2.36 3.00 3.50 8 1.40 1.86 2.31 2.90 3.36 9 1.38 1.83 2.26 2.82 3.25 10 1.37 1.81 2.23 2.76 3.17 T = 4 . 94 df = 9 What is the conclusion of the hy- pothesis test? The data provide convincing evidence of a difference between traffic flow on Friday 6 th and 13 th . µ diff = 0 x diff = 1836 − 1836 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 13 / 33

  25. Small sample inference for the mean Constructing confidence intervals using the t distribution Small sample inference for the mean 1 Introducing the t distribution Evaluating hypotheses using the t distribution Constructing confidence intervals using the t distribution Synthesis The t distribution for the difference of two means 2 Sampling distribution for the difference of two means Hypothesis testing for the difference of two means Confidence intervals for the difference of two means Recap Statistics 101 U4 - L2: t -distribution Thomas Leininger

  26. Small sample inference for the mean Constructing confidence intervals using the t distribution What is the difference? We concluded that there is a difference in the traffic flow between Friday 6 th and 13 th . Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 14 / 33

  27. Small sample inference for the mean Constructing confidence intervals using the t distribution What is the difference? We concluded that there is a difference in the traffic flow between Friday 6 th and 13 th . But it would be more interesting to find out what exactly this difference is. Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 14 / 33

  28. Small sample inference for the mean Constructing confidence intervals using the t distribution What is the difference? We concluded that there is a difference in the traffic flow between Friday 6 th and 13 th . But it would be more interesting to find out what exactly this difference is. We can use a confidence interval to estimate this difference. Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 14 / 33

  29. Small sample inference for the mean Constructing confidence intervals using the t distribution Confidence interval for a small sample mean Confidence intervals are always of the form point estimate ± ME Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 15 / 33

  30. Small sample inference for the mean Constructing confidence intervals using the t distribution Confidence interval for a small sample mean Confidence intervals are always of the form point estimate ± ME As always, ME = critical value × SE. Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 15 / 33

  31. Small sample inference for the mean Constructing confidence intervals using the t distribution Confidence interval for a small sample mean Confidence intervals are always of the form point estimate ± ME As always, ME = critical value × SE. Since small sample means follow a t distribution (and not a z distribution), the critical value is a t ⋆ (as opposed to a z ⋆ ). point estimate ± t ⋆ × SE Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 15 / 33

  32. Small sample inference for the mean Constructing confidence intervals using the t distribution Finding the critical t ( t ⋆ ) df = 9 95% CI: 95% n = 10 , df = 10 − 1 = 9 , t ⋆ is at the intersection of row df = 9 and two tail probability 0.05. 0 t* = ? one tail 0.100 0.050 0.025 0.010 0.005 two tails 0.200 0.100 0.050 0.020 0.010 df 6 1.44 1.94 2.45 3.14 3.71 7 1.41 1.89 2.36 3.00 3.50 8 1.40 1.86 2.31 2.90 3.36 9 1.38 1.83 2.26 2.82 3.25 10 1.37 1.81 2.23 2.76 3.17 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 16 / 33

  33. Small sample inference for the mean Constructing confidence intervals using the t distribution Finding the critical t ( t ⋆ ) df = 9 95% CI: 95% n = 10 , df = 10 − 1 = 9 , t ⋆ is at the intersection of row df = 9 and two tail probability 0.05. 0 t* = ? one tail 0.100 0.050 0.025 0.010 0.005 two tails 0.200 0.100 0.050 0.020 0.010 df 6 1.44 1.94 2.45 3.14 3.71 7 1.41 1.89 2.36 3.00 3.50 8 1.40 1.86 2.31 2.90 3.36 9 1.38 1.83 2.26 2.82 3.25 10 1.37 1.81 2.23 2.76 3.17 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 16 / 33

  34. Small sample inference for the mean Constructing confidence intervals using the t distribution Finding the critical t ( t ⋆ ) df = 9 95% CI: 95% n = 10 , df = 10 − 1 = 9 , t ⋆ is at the intersection of row df = 9 and two tail probability 0.05. 0 t* = ? one tail 0.100 0.050 0.025 0.010 0.005 two tails 0.200 0.100 0.050 0.020 0.010 df 6 1.44 1.94 2.45 3.14 3.71 7 1.41 1.89 2.36 3.00 3.50 8 1.40 1.86 2.31 2.90 3.36 9 1.38 1.83 2.26 2.82 3.25 10 1.37 1.81 2.23 2.76 3.17 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 16 / 33

  35. Small sample inference for the mean Constructing confidence intervals using the t distribution Finding the critical t ( t ⋆ ) df = 9 95% CI: n = 10 , df = 10 − 1 = 9 , t ⋆ is at the 95% intersection of row df = 9 and two tail probability 0.05. t = −2.26 0 t* = 2.26 one tail 0.100 0.050 0.025 0.010 0.005 two tails 0.200 0.100 0.050 0.020 0.010 df 6 1.44 1.94 2.45 3.14 3.71 7 1.41 1.89 2.36 3.00 3.50 8 1.40 1.86 2.31 2.90 3.36 9 1.38 1.83 2.26 2.82 3.25 10 1.37 1.81 2.23 2.76 3.17 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 16 / 33

  36. Small sample inference for the mean Constructing confidence intervals using the t distribution Constructing a CI for a small sample mean Question Which of the following is the correct calculation of a 95% confidence interval for the difference between the traffic flow between Friday 6 th and 13 th ? ¯ x di ff = 1836 s di ff = 1176 n = 10 SE = 372 (a) 1836 ± 1 . 96 × 372 (b) 1836 ± 2 . 26 × 372 (c) 1836 ± − 2 . 26 × 372 (d) 1836 ± 2 . 26 × 1176 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 17 / 33

  37. Small sample inference for the mean Constructing confidence intervals using the t distribution Constructing a CI for a small sample mean Question Which of the following is the correct calculation of a 95% confidence interval for the difference between the traffic flow between Friday 6 th and 13 th ? ¯ x di ff = 1836 s di ff = 1176 n = 10 SE = 372 (a) 1836 ± 1 . 96 × 372 (b) 1836 ± 2 . 26 × 372 → (995, 2677) (c) 1836 ± − 2 . 26 × 372 (d) 1836 ± 2 . 26 × 1176 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 17 / 33

  38. Small sample inference for the mean Constructing confidence intervals using the t distribution Interpreting the CI Question Which of the following is the best interpretation for the confidence in- terval we just calculated? µ di ff :6 th − 13 th = (995 , 2677) We are 95% confident that ... (a) the difference between the average number of cars on the road on Friday 6 th and 13 th is between 995 and 2,677. (b) on Friday 6 th there are 995 to 2,677 fewer cars on the road than on the Friday 13 th , on average. (c) on Friday 6 th there are 995 fewer to 2,677 more cars on the road than on the Friday 13 th , on average. (d) on Friday 13 th there are 995 to 2,677 fewer cars on the road than on the Friday 6 th , on average. Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 18 / 33

  39. Small sample inference for the mean Constructing confidence intervals using the t distribution Interpreting the CI Question Which of the following is the best interpretation for the confidence in- terval we just calculated? µ di ff :6 th − 13 th = (995 , 2677) We are 95% confident that ... (a) the difference between the average number of cars on the road on Friday 6 th and 13 th is between 995 and 2,677. (b) on Friday 6 th there are 995 to 2,677 fewer cars on the road than on the Friday 13 th , on average. (c) on Friday 6 th there are 995 fewer to 2,677 more cars on the road than on the Friday 13 th , on average. (d) on Friday 13 th there are 995 to 2,677 fewer cars on the road than on the Friday 6 th , on average. Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 18 / 33

  40. Small sample inference for the mean Synthesis Small sample inference for the mean 1 Introducing the t distribution Evaluating hypotheses using the t distribution Constructing confidence intervals using the t distribution Synthesis The t distribution for the difference of two means 2 Sampling distribution for the difference of two means Hypothesis testing for the difference of two means Confidence intervals for the difference of two means Recap Statistics 101 U4 - L2: t -distribution Thomas Leininger

  41. Small sample inference for the mean Synthesis Synthesis Does the conclusion from the hypothesis test agree with the findings of the confidence interval? Do the findings of the study suggest that people believe Friday 13 th is a day of bad luck? Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 19 / 33

  42. Small sample inference for the mean Synthesis Synthesis Does the conclusion from the hypothesis test agree with the findings of the confidence interval? Yes, the hypothesis test found a significant difference, and the CI does not contain the null value of 0. Do the findings of the study suggest that people believe Friday 13 th is a day of bad luck? Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 19 / 33

  43. Small sample inference for the mean Synthesis Synthesis Does the conclusion from the hypothesis test agree with the findings of the confidence interval? Yes, the hypothesis test found a significant difference, and the CI does not contain the null value of 0. Do the findings of the study suggest that people believe Friday 13 th is a day of bad luck? No, this is an observational study. We have just observed a significant difference between the number of cars on the road on these two days. We have not tested for people’s beliefs. Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 19 / 33

  44. Small sample inference for the mean Synthesis Recap: Inference using a small sample mean s If n < 30 , sample means follow a t distribution with SE = √ n . Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 20 / 33

  45. Small sample inference for the mean Synthesis Recap: Inference using a small sample mean s If n < 30 , sample means follow a t distribution with SE = √ n . Conditions: independence of observations n < 30 and no extreme skew Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 20 / 33

  46. Small sample inference for the mean Synthesis Recap: Inference using a small sample mean s If n < 30 , sample means follow a t distribution with SE = √ n . Conditions: independence of observations n < 30 and no extreme skew Hypothesis testing: T df = point estimate − null value , where df = n − 1 SE Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 20 / 33

  47. Small sample inference for the mean Synthesis Recap: Inference using a small sample mean s If n < 30 , sample means follow a t distribution with SE = √ n . Conditions: independence of observations n < 30 and no extreme skew Hypothesis testing: T df = point estimate − null value , where df = n − 1 SE Confidence interval: point estimate ± t ⋆ df × SE Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 20 / 33

  48. Small sample inference for the mean Synthesis Recap: Inference using a small sample mean s If n < 30 , sample means follow a t distribution with SE = √ n . Conditions: independence of observations n < 30 and no extreme skew Hypothesis testing: T df = point estimate − null value , where df = n − 1 SE Confidence interval: point estimate ± t ⋆ df × SE Note: The example we used was for paired means (difference between dependent groups). We took the difference between the observations and used only these differences (one sample) in our analysis, therefore the mechanics are the same as when we are working with just one sample. Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 20 / 33

  49. The t distribution for the difference of two means Small sample inference for the mean 1 Introducing the t distribution Evaluating hypotheses using the t distribution Constructing confidence intervals using the t distribution Synthesis The t distribution for the difference of two means 2 Sampling distribution for the difference of two means Hypothesis testing for the difference of two means Confidence intervals for the difference of two means Recap Statistics 101 U4 - L2: t -distribution Thomas Leininger

  50. The t distribution for the difference of two means Diamonds Weights of diamonds are measured in carats. 1 carat = 100 points, 0.99 carats = 99 points, etc. The difference between the size of a 0.99 carat diamond and a 1 carat diamond is undetectable to the naked human eye, but the price of a 1 carat diamond tends to be much higher than the price of a 0.99 diamond. We are going to test to see if there is a difference between the average prices of 0.99 and 1 carat diamonds. In order to be able to compare equivalent units, we divide the prices of 0.99 carat diamonds by 99 and 1 carat diamonds by 100, and compare the average point prices. Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 21 / 33

  51. The t distribution for the difference of two means Data 80 70 60 50 40 30 20 carat = 0.99 carat = 1 0.99 carat 1 carat pt99 pt100 x ¯ 44.50 53.43 s 13.32 12.22 n 23 30 These data are a random sample from the diamonds data set in ggplot2 R package. Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 22 / 33

  52. The t distribution for the difference of two means Parameter and point estimate Parameter of interest: Average difference between the point prices of all 0.99 carat and 1 carat diamonds. µ pt 99 − µ pt 100 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 23 / 33

  53. The t distribution for the difference of two means Parameter and point estimate Parameter of interest: Average difference between the point prices of all 0.99 carat and 1 carat diamonds. µ pt 99 − µ pt 100 Point estimate: Average difference between the point prices of sampled 0.99 carat and 1 carat diamonds. x pt 99 − ¯ ¯ x pt 100 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 23 / 33

  54. The t distribution for the difference of two means Hypotheses Question Which of the following is the correct set of hypotheses for testing if the average point price of 1 carat diamonds ( pt 100 ) is higher than the average point price of 0.99 carat diamonds ( pt 99 )? Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 24 / 33

  55. The t distribution for the difference of two means Hypotheses Question Which of the following is the correct set of hypotheses for testing if the average point price of 1 carat diamonds ( pt 100 ) is higher than the average point price of 0.99 carat diamonds ( pt 99 )? (a) H 0 : µ pt 99 = µ pt 100 H A : µ pt 99 � µ pt 100 (b) H 0 : µ pt 99 = µ pt 100 H A : µ pt 99 > µ pt 100 (c) H 0 : µ pt 99 = µ pt 100 H A : µ pt 99 < µ pt 100 (d) H 0 : ¯ x pt 99 = ¯ x pt 100 H A : ¯ x pt 99 < ¯ x pt 100 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 24 / 33

  56. The t distribution for the difference of two means Conditions Question What conditions need to be satisfied in order to conduct this hypothesis test using theoretical methods? Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 25 / 33

  57. The t distribution for the difference of two means Conditions Question What conditions need to be satisfied in order to conduct this hypothesis test using theoretical methods? Point price of one 0.99 carat diamond in the sample should be independent of another, and the point price of one 1 carat diamond should independent of another as well. Point prices of 0.99 carat and 1 carat diamonds in the sample should be independent. Distributions of point prices of 0.99 and 1 carat diamonds should not be extremely skewed. Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 25 / 33

  58. The t distribution for the difference of two means Sampling distribution for the difference of two means Small sample inference for the mean 1 Introducing the t distribution Evaluating hypotheses using the t distribution Constructing confidence intervals using the t distribution Synthesis The t distribution for the difference of two means 2 Sampling distribution for the difference of two means Hypothesis testing for the difference of two means Confidence intervals for the difference of two means Recap Statistics 101 U4 - L2: t -distribution Thomas Leininger

  59. The t distribution for the difference of two means Sampling distribution for the difference of two means Test statistic Test statistic for inference on the difference of two small sample means The test statistic for inference on the difference of two small sample means ( n 1 < 30 and/or n 2 < 30 ) mean is the T statistic. T df = point estimate − null value SE where � s 2 s 2 1 2 SE = and df = min ( n 1 − 1 , n 2 − 1) + n 1 n 2 Note: The calculation of the df is actually much more complicated. For simplicity we’ll use the above formula to estimate the true df when conducting the analysis by hand. Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 26 / 33

  60. The t distribution for the difference of two means Hypothesis testing for the difference of two means Small sample inference for the mean 1 Introducing the t distribution Evaluating hypotheses using the t distribution Constructing confidence intervals using the t distribution Synthesis The t distribution for the difference of two means 2 Sampling distribution for the difference of two means Hypothesis testing for the difference of two means Confidence intervals for the difference of two means Recap Statistics 101 U4 - L2: t -distribution Thomas Leininger

  61. The t distribution for the difference of two means Hypothesis testing for the difference of two means Test statistic (cont.) 0.99 carat 1 carat pt99 pt100 x ¯ 44.50 53.43 s 13.32 12.22 23 30 n in context... Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 27 / 33

  62. The t distribution for the difference of two means Hypothesis testing for the difference of two means Test statistic (cont.) 0.99 carat 1 carat pt99 pt100 x ¯ 44.50 53.43 s 13.32 12.22 23 30 n in context... point estimate − null value T = SE Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 27 / 33

  63. The t distribution for the difference of two means Hypothesis testing for the difference of two means Test statistic (cont.) 0.99 carat 1 carat pt99 pt100 x ¯ 44.50 53.43 s 13.32 12.22 23 30 n in context... point estimate − null value T = SE (44 . 50 − 53 . 43) − 0 = � 13 . 32 2 + 12 . 22 2 23 30 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 27 / 33

  64. The t distribution for the difference of two means Hypothesis testing for the difference of two means Test statistic (cont.) 0.99 carat 1 carat pt99 pt100 x ¯ 44.50 53.43 s 13.32 12.22 23 30 n in context... point estimate − null value T = SE (44 . 50 − 53 . 43) − 0 = � 13 . 32 2 + 12 . 22 2 23 30 − 8 . 93 = 3 . 56 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 27 / 33

  65. The t distribution for the difference of two means Hypothesis testing for the difference of two means Test statistic (cont.) 0.99 carat 1 carat pt99 pt100 x ¯ 44.50 53.43 s 13.32 12.22 23 30 n in context... point estimate − null value T = SE (44 . 50 − 53 . 43) − 0 = � 13 . 32 2 + 12 . 22 2 23 30 − 8 . 93 = 3 . 56 − 2 . 508 = Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 27 / 33

  66. The t distribution for the difference of two means Hypothesis testing for the difference of two means Test statistic (cont.) Question Which of the following is the correct df for this hypothesis test? (a) 22 (b) 23 (c) 30 (d) 29 (e) 52 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 28 / 33

  67. The t distribution for the difference of two means Hypothesis testing for the difference of two means Test statistic (cont.) Question Which of the following is the correct df for this hypothesis test? (a) 22 → df = min ( n pt 99 − 1 , n pt 100 − 1) = min (23 − 1 , 30 − 1) (b) 23 = min (22 , 29) = 22 (c) 30 (d) 29 (e) 52 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 28 / 33

  68. The t distribution for the difference of two means Hypothesis testing for the difference of two means p-value Question Which of the following is the correct p-value for this hypothesis test? T = − 2 . 508 df = 22 one tail 0.100 0.050 0.025 0.010 two tails 0.200 0.100 0.050 0.020 (a) between 0.005 and 0.01 df 21 1.32 1.72 2.08 2.52 (b) between 0.01 and 0.025 22 1.32 1.72 2.07 2.51 23 1.32 1.71 2.07 2.50 (c) between 0.02 and 0.05 24 1.32 1.71 2.06 2.49 (d) between 0.01 and 0.02 25 1.32 1.71 2.06 2.49 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 29 / 33

  69. The t distribution for the difference of two means Hypothesis testing for the difference of two means p-value Question Which of the following is the correct p-value for this hypothesis test? T = − 2 . 508 df = 22 one tail 0.100 0.050 0.025 0.010 two tails 0.200 0.100 0.050 0.020 (a) between 0.005 and 0.01 df 21 1.32 1.72 2.08 2.52 (b) between 0.01 and 0.025 22 1.32 1.72 2.07 2.51 23 1.32 1.71 2.07 2.50 (c) between 0.02 and 0.05 24 1.32 1.71 2.06 2.49 (d) between 0.01 and 0.02 25 1.32 1.71 2.06 2.49 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 29 / 33

  70. The t distribution for the difference of two means Hypothesis testing for the difference of two means Synthesis What is the conclusion of the hypothesis test? How (if at all) would this conclusion change your behavior if you went diamond shopping? Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 30 / 33

  71. The t distribution for the difference of two means Hypothesis testing for the difference of two means Synthesis What is the conclusion of the hypothesis test? How (if at all) would this conclusion change your behavior if you went diamond shopping? p-value is small so reject H 0 . The data provide convincing evidence to suggest that the point price of 0.99 carat diamonds is lower than the point price of 1 carat diamonds. Maybe buy a 0.99 carat diamond? It looks like a 1 carat, but is significantly cheaper. Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 30 / 33

  72. The t distribution for the difference of two means Confidence intervals for the difference of two means Small sample inference for the mean 1 Introducing the t distribution Evaluating hypotheses using the t distribution Constructing confidence intervals using the t distribution Synthesis The t distribution for the difference of two means 2 Sampling distribution for the difference of two means Hypothesis testing for the difference of two means Confidence intervals for the difference of two means Recap Statistics 101 U4 - L2: t -distribution Thomas Leininger

  73. The t distribution for the difference of two means Confidence intervals for the difference of two means Critical value Question What is the appropriate t ⋆ for a 90% confidence interval for the average difference between the point prices of 0.99 and 1 carat diamonds? one tail 0.100 0.050 0.025 0.010 0.005 two tails 0.200 0.100 0.050 0.020 0.010 df 21 1.32 1.72 2.08 2.52 2.83 22 1.32 1.72 2.07 2.51 2.82 23 1.32 1.71 2.07 2.50 2.81 24 1.32 1.71 2.06 2.49 2.80 25 1.32 1.71 2.06 2.49 2.79 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 31 / 33

  74. The t distribution for the difference of two means Confidence intervals for the difference of two means Critical value Question What is the appropriate t ⋆ for a 90% confidence interval for the average difference between the point prices of 0.99 and 1 carat diamonds? one tail 0.100 0.050 0.025 0.010 0.005 two tails 0.200 0.100 0.050 0.020 0.010 df 21 1.32 1.72 2.08 2.52 2.83 22 1.32 1.72 2.07 2.51 2.82 23 1.32 1.71 2.07 2.50 2.81 24 1.32 1.71 2.06 2.49 2.80 25 1.32 1.71 2.06 2.49 2.79 Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 31 / 33

  75. The t distribution for the difference of two means Confidence intervals for the difference of two means Confidence interval Calculate the interval, and interpret it in context. Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 32 / 33

  76. The t distribution for the difference of two means Confidence intervals for the difference of two means Confidence interval Calculate the interval, and interpret it in context. point estimate ± ME Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 32 / 33

  77. The t distribution for the difference of two means Confidence intervals for the difference of two means Confidence interval Calculate the interval, and interpret it in context. point estimate ± ME x pt 1 ) ± t ⋆ (¯ x pt 99 − ¯ df × SE (44 . 50 − 53 . 43) ± 1 . 72 × 3 . 56 = Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 32 / 33

  78. The t distribution for the difference of two means Confidence intervals for the difference of two means Confidence interval Calculate the interval, and interpret it in context. point estimate ± ME x pt 1 ) ± t ⋆ (¯ x pt 99 − ¯ df × SE (44 . 50 − 53 . 43) ± 1 . 72 × 3 . 56 = − 8 . 93 ± 6 . 12 = Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 32 / 33

  79. The t distribution for the difference of two means Confidence intervals for the difference of two means Confidence interval Calculate the interval, and interpret it in context. point estimate ± ME x pt 1 ) ± t ⋆ (¯ x pt 99 − ¯ df × SE (44 . 50 − 53 . 43) ± 1 . 72 × 3 . 56 = − 8 . 93 ± 6 . 12 = ( − 15 . 05 , − 2 . 81) = Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 32 / 33

  80. The t distribution for the difference of two means Confidence intervals for the difference of two means Confidence interval Calculate the interval, and interpret it in context. point estimate ± ME x pt 1 ) ± t ⋆ (¯ x pt 99 − ¯ df × SE (44 . 50 − 53 . 43) ± 1 . 72 × 3 . 56 = − 8 . 93 ± 6 . 12 = ( − 15 . 05 , − 2 . 81) = We are 90% confident that the average point price of a 0.99 carat diamond is $15.05 to $2.81 lower than the average point price of a 1 carat diamond. Statistics 101 (Thomas Leininger) U4 - L2: t -distribution June 5, 2013 32 / 33

  81. The t distribution for the difference of two means Recap Small sample inference for the mean 1 Introducing the t distribution Evaluating hypotheses using the t distribution Constructing confidence intervals using the t distribution Synthesis The t distribution for the difference of two means 2 Sampling distribution for the difference of two means Hypothesis testing for the difference of two means Confidence intervals for the difference of two means Recap Statistics 101 U4 - L2: t -distribution Thomas Leininger

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