MQ-STAT2170-6180-Exam-S1-2022

This question paper must be returned. Candidates are not permitted to remove any part of it from the examination room. SEAT NUMBER: ……….… ROOM:..………………. FAMILY NAME:………… ..... …………………………. OTHER NAMES:……..…….…………………..…….. STUDENT NUMBER:…..…….………..…………….. C opyright © Macquarie University. Copying or distribution of part or all of the contents in any form is prohibited. FORMAL EXAMINATION PERIOD: SESSION 1, JUNE 2022 Unit Code: STAT2170 & STAT6180 Unit Name: Applied Statistics Duration of Exam (including reading time if applicable) : Two (2) hours and ten (10) minutes Total No. of Questions: Four (4) Total No. of Pages (including this cover sheet) : 21 GENERAL INSTRUCTIONS TO STUDENTS: • Students are required to follow directions given by the Final Examination Supervisor and must refrain from communicating in any way with another student once they have entered the final examination venue. • Students may not write or mark the exam materials in any way during reading time. • Students may only access authorised materials during this examination. A list of authorised material is available on this cover sheet. • All watches must be removed and placed at the top of the exam desk and must remain there for the duration of the exam. All alarms, notifications and alerts must be switched off. • Students are not permitted to leave the exam room during the first hour (excluding reading time) and during the last 15 minutes of the examination. • If it is alleged you have breached these rules at any time during the examination, the matter may be reported to a University Discipline Committee for determination. EXAMINATION INSTRUCTIONS: • Answer all four (4) questions. • The total mark for this paper is 65 . • The value of each question is shown in brackets next to the question number and in the table below. • All answers should be written in the spaces provided . If there is insufficient space for your answer, you may use the reverse of the adjacent page and label your answers appropriately. • Statistical tables are provided at the end of the paper. You may use these if relevant. • Unless otherwise indicated, all statistical tests should be carried out using a 5% significance level. AIDS AND MATERIALS PERMITTED/NOT PERMITTED: Dictionaries: No dictionaries permitted Calculators: Non-programmable calculators with no text storage/retrieval capacity permitted Other: Open book – unrestricted hardcopy reference materials permitted Q1 Q2 Q3 Q4 Total 18 17 17 13 65

Question 1 [18 marks] In a study of the e ↵ ects of marijuana on mice, ordinary mice were divided in four groups. The first group of mice ( VEH ) received a vehicle only, which is a shot with the same inactive ingredients as the other mice but no THC (the active ingredient in marijuana). The three other groups of mice received di ↵ erent dosages of THC (0.3, 1 and 3 mg/kg respectively). The inves- tigators then measured the locomotor activity of the mice by placing each mouse in a box lined with photocells to measure the total distance covered by that mouse, and the results are reported as the percentage movement relative to the untreated mice. a) [2 marks] Comment on two interesting features of the mice THC data based on the boxplot below. VEH 0.3 1 3 40 80 120 Boxplots of Mice THC data THC dosage Activity % b) [3 marks] Suppose you wish to perform a one-way ANOVA for the Mice THC data. Use the descriptive statistics below (from RStudio) to complete the ANOVA table on the next page by filling in the missing entries (missing entries are underlined). Show your full working in the space below. str (mice_pot_new) # ’data.frame’: 46 obs. of 2 variables: # $ group : Factor w/ 4 levels "VEH","0.3","1",..: 2 2 2 2 2 2 2 2 2 3 ... # $ percent_of_act: num 98.8 116.4 132.1 84.1 148.9 ... tapply (percent_of_act, group, mean) # mean of activity by dosage group 2 The box plot look constantly, except for the first group of mice has much lower" of activity -

c) [5 marks] Based on the results in your completed ANOVA table above in part b), test if there is any di ↵ erence in activities across the four dosage groups. Remember to state the null and alternative hypotheses, give the test statistic value and associated degrees of freedom, the P-value or range of it, and draw your conclusion ( assuming that all test assumptions are ok for this part ). 4 Ho: no difference between the four dosage groups. H1: There is difference between at least two of the dosage groups, at least one group mean is different F-statistic is 3.13 with 3 and 42 df. look through F-statistic table, I critical value will be around 2.6 - 3.13) so that p valueless than 0.05 S -

d) [8 marks] If someone wants to find out whether there is any di ↵ erence in average activity between the first three dosages ( VEH , .3 mg/kg, 1 mg/kg) as a whole and the highest dosage (3 mg/kg), form an appropriate contrast to evaluate and test the di ↵ erence. Remember to state null and alternative hypotheses, give the test statistic value and associated degrees of freedom, the evidence for rejecting or not rejecting the null hypothesis with a conclusion. tapply (percent_of_act, group, mean) # VEH 0.3 1 3 # 100.00000 97.32250 99.05235 70.66787 5 Ho: the combined mean of the first three groups equal the mean of the last group Ho-MrEn + P0.3 + 4 = 3M3 Ha: difference in activity between the combined first three dosage groups and the highest dosage groups Hea MrEn + Po.3 + 4E 3M (C) observed contrast MrEc P0.3 +*-3M3 =100 97.3225 99.05235 - 3570.66787 = 85.37139 SECC) = saut [(( 11352 x 25.323952/y5) + ((11352 + 31.4565422/9) +(113(2 x 26.257822/12) - (12429.0535/10)] T 2 /SEK) = with 42dt. check if I statistic is better than critical value or not then state the hypothesis.

# Call: # lm(formula = Y ~ X + I(X^2), data = examQ2_data) # # Residuals: # Min 1Q Median 3Q Max # -114.684 -23.235 6.093 25.091 130.768 # # Coefficients: # Estimate Std. Error t value Pr(>|t|) # (Intercept) -24.0856 17.7572 -1.356 0.181 # X 4.8533 3.2849 1.477 0.146 # I(X^2) 0.8582 0.1271 6.754 1.75e-08 *** # --- # Signif. codes: 0 ’***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1 # # Residual standard error: 43.94 on 48 degrees of freedom # Multiple R-squared: 0.9557,Adjusted R-squared: 0.9539 # F-statistic: 518 on 2 and 48 DF, p-value: < 2.2e-16 a) [2 marks] Write down the two fitted regression equations corresponding with the two models, i.e., writing ˆ Y explicitly as two functions of the predictor variable X with the appropriate coe ffi cients. b) [2 marks] What is the predicted value ˆ Y for Y when X = 10 for the two models? c) [2 marks] Given that Y = 151 . 429 when X = 10 in the actual dataset, what are the corresponding residual values for the two models when X = 10? 7 The residuals are the di ff erence between the observed and predicted values. For X=10, the observed value is Y=151.429, so: For model Q2_M1, the residual is: 151.429 - 151.38 = 0.049 For model Q2_M2, the residual is: 151.429 - 111.9 = 39.529 Q2 -M, y=- 111.692 + 26.300X Q2 M2 4 = - 24.0856 + 4.853310 + 0.8582182 Q2 - MI 4 = -111.692 26.308 x 10 151.30 &2 - M2 4 = - 24.0856 + 4.853310 + 0.8582182-111.3

d) [6 marks] Given that the data set contains 51 observations, the average ¯ X = 12 . 5, the sample standard deviation s X = 7 . 433 and the residual standard error in the information above, (i) [3 marks] Find the 95% confidence interval for the slope in the linear regression model. The following information may be useful: qt ( .950 , df = 49 : 51 ) # [1] 1.676551 1.675905 1.675285 qt ( .975 , df = 49 : 51 ) # [1] 2.009575 2.008559 2.007584 (ii) [3 marks] Find the 95% prediction interval at X = 10 in the linear regression model. e) [2 marks] What are the coe ffi cients of determination, i.e. R 2 and adj R 2 , for both models, and what do they tell you about the variation in the data? 8 The 95% confidence interval for the slope in the linear regression model is given by the formula β ̂ ± t*SE( β ̂ ). From the summary of Q2_M1, β ̂ = 26.308 and SE( β ̂ ) = 1.155. The t- value with 49 degrees of freedom for a 95% confidence level is 2.009575. Therefore, the 95% confidence interval for the slope is: 26.308 ± 2.009575*1.155 = (23.34, 29.28) We can calculate Σ (Xi - X¯)^2 using the formula for variance (since it is the sum of squared deviations divided by n-1). This is equal to (n-1)*sX^2. Therefore, Σ (Xi - X¯)^2 = (51 - 1) * 7.433^2 = 2213.98. Then, we can compute the standard error of prediction SE(Yˆ) as follows: SE(Yˆ) = sqrt[RSE^2 * (1 + 1/n + (X - X¯)^2 / Σ (Xi - X¯)^2)] = sqrt[60.73^2 * (1 + 1/51 + (10 - 12.5)^2 / 2213.98)] Now that we have SE(Yˆ), we can calculate the prediction interval. The t-value with 49 degrees of freedom for a 95% confidence level is 2.009575. Therefore, the 95% prediction interval at X = 10 is: Yˆ ± t * SE(Yˆ) = 151.38 ± 2.009575 * SE(Yˆ) The R^2 value is a measure of how much of the variation in the dependent variable (Y) is explained by the predictor variable(s) (X). A higher R^2 means a larger proportion of the variance is explained. The adjusted R^2 also takes into account the number of predictors in the model, adjusting the R^2 downwards for each additional predictor to penalize for model complexity. In this case, both R^2 and adj R^2 are higher for model Q2_M2, suggesting that the quadratic model explains more of the variation in the data compared to the linear model. RSE) Residual standard error is 60.75) Q2-M1: R = 0.9136, ad R2 = 0.9119. G2-M: R2 = 0.9057, adjust R2 = 0.9539

Question 3 [17 marks] In a survey of 500 towns, the relation of the frequency of biking to work and the frequency of smoking on the one hand to the frequency of heart disease on the other was investigated. head (heart) # biking smoking heart.disease # 1 30.801246 10.896608 11.769423 # 2 65.129215 2.219563 2.854081 # 3 1.959665 17.588331 17.177803 # 4 44.800196 2.802559 6.816647 # 5 69.428454 15.974505 4.062224 # 6 54.403626 29.333176 9.550046 str (heart) # ’data.frame’: 498 obs. of 3 variables: # $ biking : num 30.8 65.13 1.96 44.8 69.43 ... # $ smoking : num 10.9 2.22 17.59 2.8 15.97 ... # $ heart.disease: num 11.77 2.85 17.18 6.82 4.06 ... plot (heart) biking 0 15 30 0 20 40 60 0 5 10 15 20 25 30 smoking 0 40 0 5 10 15 20 0 10 20 heart.disease In this question, we will use a multiple regression model with two predictor variables. a) [2 marks] The summary output for a proposed multiple regression model with a quadratic term is shown below: Heart_mod1 = lm (heart.disease ~ biking + smoking + I (smoking ^ 2 ), data = heart) summary (Heart_mod1) # # Call: # lm(formula = heart.disease ~ biking + smoking + I(smoking^2), # data = heart) # # Residuals: 10

# Min 1Q Median 3Q Max # -2.17529 -0.44416 0.03145 0.44166 1.95840 # # Coefficients: # Estimate Std. Error t value Pr(>|t|) # (Intercept) 15.0503322 0.1079806 139.380 <2e-16 *** # biking -0.2000727 0.0013677 -146.282 <2e-16 *** # smoking 0.1657528 0.0143063 11.586 <2e-16 *** # I(smoking^2) 0.0004114 0.0004533 0.908 0.365 # --- # Signif. codes: 0 ’***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1 # # Residual standard error: 0.6541 on 494 degrees of freedom # Multiple R-squared: 0.9797,Adjusted R-squared: 0.9795 # F-statistic: 7928 on 3 and 494 DF, p-value: < 2.2e-16 Is the model adequate, or do we need to conduct further analysis? b) [2 marks] The second statistical model that we then use is given by Y = β 0 + β 1 X 1 + β 2 X 2 + " where X 1 , X 2 , Y and " are random variables. Explain the meaning of the various regression coe ffi cients. c) [2 marks] The summary output for the second multiple regression model is shown below: Heart_mod2 = lm (heart.disease ~ biking + smoking, data = heart) summary (Heart_mod2) # # Call: # lm(formula = heart.disease ~ biking + smoking, data = heart) # # Residuals: # Min 1Q Median 3Q Max # -2.1789 -0.4463 0.0362 0.4422 1.9331 # # Coefficients: # Estimate Std. Error t value Pr(>|t|) # (Intercept) 14.984658 0.080137 186.99 <2e-16 *** 11 β 0 (the intercept): It represents the expected mean value of the response variable (Y) when all predictor variables (X1 and X2) are zero. It is the value of Y when there is no biking and no smoking. β 1 (coe ffi cient for X1): It represents the change in the expected mean value of Y for a one-unit increase in the predictor variable X1 (biking), assuming X2 (smoking) is held constant. In other words, it measures the linear e ff ect of biking on the mean value of Y. β 2 (coe ffi cient for X2): It represents the change in the expected mean value of Y for a one-unit increase in the predictor variable X2 (smoking), assuming X1 (biking) is held constant. It measures the linear e ff ect of smoking on the mean value of Y. ε (the error term): It represents the random variability in the response variable Y that is not explained by the predictor variables X1 and X2. It includes all other factors that a ff ect Y but are not accounted for in the model. Tocheck if the model is adequate, (PLZt-16) -> the coefficiant for the intercept, biking, and smoking are all highly significant. However, coef of quadric term (p = 0.365) - not signif -> quadricterm may not be necessary R2 0.975 -> model explain 97 : 95% of the variation -> model is look good however further analysis if quadricterm ICsmoking ? ) is necessary.

To validate the second linear model, there are various assumptions that we have to verify. What do we have to validate, and what can you validate? e) [4 marks] The sequential ANOVA tables for the multiple linear regression model are given below. heart_anova1 = anova ( lm (heart.disease ~ biking + smoking, data = heart)) heart_anova2 = anova ( lm (heart.disease ~ smoking + biking, data = heart)) heart_anova1 # Analysis of Variance Table # # Response: heart.disease # Df Sum Sq Mean Sq F value Pr(>F) # biking 1 9090.6 9090.6 21251.7 < 2.2e-16 *** # smoking 1 1086.0 1086.0 2538.8 < 2.2e-16 *** # Residuals 495 211.7 0.4 # --- # Signif. codes: 0 ’***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1 heart_anova2 # Analysis of Variance Table # # Response: heart.disease # Df Sum Sq Mean Sq F value Pr(>F) # smoking 1 992.7 992.7 2320.8 < 2.2e-16 *** # biking 1 9183.8 9183.8 21469.7 < 2.2e-16 *** # Residuals 495 211.7 0.4 # --- # Signif. codes: 0 ’***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1 13 Linearity: We assume that the relationship between the predictors (biking and smoking) and the response variable (heart disease) is linear. We can assess this assumption by examining the plot of standardized residuals vs. fitted values (the first diagnostic plot). If the points in the plot are randomly scattered around the horizontal line at zero, it suggests that the linearity assumption holds. Independence of residuals: We assume that the residuals (the di ff erences between observed and predicted values) are independent of each other. We can check this assumption by examining the plot of residuals vs. the observation index (the second diagnostic plot). If there is no discernible pattern or trend in the plot, it indicates that the independence assumption is reasonable. Homoscedasticity: We assume that the variance of the residuals is constant across all levels of the predictors. We can assess this assumption by examining the plot of residuals vs. the predictor variables (biking and smoking). If the spread of the residuals is relatively constant and there is no discernible funnel shape or pattern, it suggests that the homoscedasticity assumption is met. Normality of residuals: We assume that the residuals follow a normal distribution. This assumption is di ffi cult to directly assess from the provided diagnostic plots. However, if the sample size is large (in this case, 500 towns), the Central Limit Theorem suggests that the residuals are likely to be approximately normally distributed. Based on the diagnostic plots provided, we can validate the linearity assumption to some extent as the first plot shows no clear patterns or deviations from linearity. The second plot of residuals vs. observation index suggests that there is no significant violation of independence of residuals. However, we cannot directly assess the homoscedasticity assumption or the normality of residuals from the plots given.

What is the di ↵ erence between these two tables? What proportion of the total sum of squares is explained by the variable biking alone, independent of the second variable? And what proportion by the variable smoking independent of biking ? f) [3 marks] Suppose now that you decide to do a simple linear regression with the obvious response variable heart.disease and predictor variable biking . Add the missing entries in the table below. Source d.f. S.S. M.S. F-value biking Residual Total 14 In heart_anova1, the model includes biking as the first predictor and smoking as the second predictor. It shows the analysis of variance for the contributions of biking and smoking to the total sum of squares. In heart_anova2, the model includes smoking as the first predictor and biking as the second predictor. It shows the analysis of variance for the contributions of smoking and biking to the total sum of squares. To determine the proportion of the total sum of squares explained by each variable independently, we look at the "Mean Sq" column in each ANOVA table. For heart_anova1: Proportion explained by biking alone: Mean Sq for biking / (Mean Sq for biking + Mean Sq for smoking + Mean Sq for Residuals) = 9090.6 / (9090.6 + 1086.0 + 211.7) For heart_anova2: Proportion explained by smoking alone: Mean Sq for smoking / (Mean Sq for smoking + Mean Sq for biking + Mean Sq for Residuals) = 992.7 / (992.7 + 9183.8 + 211.7)

b) [1 mark] What information can you determine from the interaction plot in Figure 1 on p.15? c) [3 marks] Two-Way ANOVA models with and without interactions are fitted and the re- sulting ANOVA tables were provided below. The data is analysed in R output below. # Analysis of Variance Table # # Response: proficiency # Df Sum Sq Mean Sq F value Pr(>F) # graduation 2 152.33 76.17 14.5851 0.001501 ** # method 2 849.33 424.67 81.3191 1.731e-06 *** # graduation:method 4 17.33 4.33 0.8298 0.538702 # Residuals 9 47.00 5.22 # --- # Signif. codes: 0 ’***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1 # Analysis of Variance Table # # Response: proficiency # Df Sum Sq Mean Sq F value Pr(>F) # graduation 2 152.33 76.17 15.391 0.0003734 *** # method 2 849.33 424.67 85.814 3.234e-08 *** # Residuals 13 64.33 4.95 # --- # Signif. codes: 0 ’***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1 Choose and justify the most appropriate model in this scenario that explains the data. Then also state the overall conclusions about the e ↵ ect of training methods and time elapsed from university graduation. 16 To choose the most appropriate model, we can compare the significance of the interaction term in the ANOVA table. In the model with interactions, the p-value for the interaction term (graduation:method) is 0.538702, which is not significant. This suggests that there is no evidence of interaction e ff ects between graduation and method. Therefore, the most appropriate model in this scenario would be the model without interactions. This model indicates that both training method and graduation have significant e ff ects on proficiency scores, independent of each other. Overall, both training methods and time elapsed from university graduation have significant e ff ects on auditor proficiency. However, the specific nature of these e ff ects and their interactions are not explicitly captured in the analysis provided. The graduation factor has a significant e ff ect on proficiency scores (p < 0.001), indicating that the time elapsed since university graduation has an impact on auditor proficiency. The training method factor also has a significant e ff ect on proficiency scores (p < 0.001), suggesting that di ff erent training methods lead to di ff erent levels of proficiency among auditors. The absence of a significant interaction between graduation and method suggests that the e ff ect of graduation on proficiency does not depend on the training method employed, and vice versa.

d) [1 mark] Given the design of the data, are the Tukey HDS multiple comparisons appropriate here? Justify your answer and and if necessary, suggest an alternative technique. e) [6 marks] Using the Tukey HDS multiple comparisons, what are your conclusions about the training method on the proficiency scores? If any comparisons are significant, state their 95% CI if possible and explain their meaning. The following output may be useful. # Tukey multiple comparisons of means # 95% family-wise confidence level # # Fit: aov(formula = lm.audit.int) # # $graduation # diff lwr upr p adj # 2-1 -1.666667 -5.350359 2.017026 0.4488408 # 3-1 -6.833333 -10.517026 -3.149641 0.0015046 # 3-2 -5.166667 -8.850359 -1.482974 0.0089043 # # $method # diff lwr upr p adj # B-A 4.666667 0.9829745 8.350359 0.0157656 # C-A 16.333333 12.6496411 20.017026 0.0000016 # C-B 11.666667 7.9829745 15.350359 0.0000263 # # $‘graduation:method‘ # diff lwr upr p adj # 2:A-1:A 1.421085e-14 -9.04045793 9.0404579 1.0000000 # 3:A-1:A -8.500000e+00 -17.54045793 0.5404579 0.0689359 # 1:B-1:A 5.000000e+00 -4.04045793 14.0404579 0.4791591 # 2:B-1:A 2.000000e+00 -7.04045793 11.0404579 0.9891520 # 3:B-1:A -1.500000e+00 -10.54045793 7.5404579 0.9983230 # 1:C-1:A 1.600000e+01 6.95954207 25.0404579 0.0011954 # 2:C-1:A 1.400000e+01 4.95954207 23.0404579 0.0031741 # 3:C-1:A 1.050000e+01 1.45954207 19.5404579 0.0212665 # 3:A-2:A -8.500000e+00 -17.54045793 0.5404579 0.0689359 # 1:B-2:A 5.000000e+00 -4.04045793 14.0404579 0.4791591 # 2:B-2:A 2.000000e+00 -7.04045793 11.0404579 0.9891520 # 3:B-2:A -1.500000e+00 -10.54045793 7.5404579 0.9983230 # 1:C-2:A 1.600000e+01 6.95954207 25.0404579 0.0011954 # 2:C-2:A 1.400000e+01 4.95954207 23.0404579 0.0031741 # 3:C-2:A 1.050000e+01 1.45954207 19.5404579 0.0212665 # 1:B-3:A 1.350000e+01 4.45954207 22.5404579 0.0041032 # 2:B-3:A 1.050000e+01 1.45954207 19.5404579 0.0212665 # 3:B-3:A 7.000000e+00 -2.04045793 16.0404579 0.1668333 # 1:C-3:A 2.450000e+01 15.45954207 33.5404579 0.0000404 17 The Tukey Honestly Significant Di ff erence (HSD) test is typically used for pairwise comparisons in an analysis of variance (ANOVA) when the assumption of equal variances holds. However, in this scenario, the design of the data does not meet the assumptions required for the Tukey HSD test.The Tukey HSD assumes that the data come from a balanced design, where the sample sizes are equal across all treatment groups. In this case, the design is not fully balanced because the auditors were grouped into blocks based on time elapsed since university graduation. Each graduation group consists of auditors with di ff erent sample sizes. Consequently, the assumption of equal sample sizes across all treatment groups is violated.

Standard normal probabilities 0 z P ( Z  z ) Z ⇠ N (0 , 1) Key: Table entry for z is the area under the standard normal curve to the left of z , that is, P ( Z  z ). z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 - 2 . 9 0.0019 0.0018 0.0018 0.0017 0.0016 0.0016 0.0015 0.0015 0.0014 0.0014 - 2 . 8 0.0026 0.0025 0.0024 0.0023 0.0023 0.0022 0.0021 0.0021 0.0020 0.0019 - 2 . 7 0.0035 0.0034 0.0033 0.0032 0.0031 0.0030 0.0029 0.0028 0.0027 0.0026 - 2 . 6 0.0047 0.0045 0.0044 0.0043 0.0041 0.0040 0.0039 0.0038 0.0037 0.0036 - 2 . 5 0.0062 0.0060 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051 0.0049 0.0048 - 2 . 4 0.0082 0.0080 0.0078 0.0075 0.0073 0.0071 0.0069 0.0068 0.0066 0.0064 - 2 . 3 0.0107 0.0104 0.0102 0.0099 0.0096 0.0094 0.0091 0.0089 0.0087 0.0084 - 2 . 2 0.0139 0.0136 0.0132 0.0129 0.0125 0.0122 0.0119 0.0116 0.0113 0.0110 - 2 . 1 0.0179 0.0174 0.0170 0.0166 0.0162 0.0158 0.0154 0.0150 0.0146 0.0143 - 2 . 0 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183 - 1 . 9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.0233 - 1 . 8 0.0359 0.0351 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307 0.0301 0.0294 - 1 . 7 0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384 0.0375 0.0367 - 1 . 6 0.0548 0.0537 0.0526 0.0516 0.0505 0.0495 0.0485 0.0475 0.0465 0.0455 - 1 . 5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559 - 1 . 4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0721 0.0708 0.0694 0.0681 - 1 . 3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823 - 1 . 2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 0.0985 - 1 . 1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.1230 0.1210 0.1190 0.1170 - 1 . 0 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379 - 0 . 9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711 0.1685 0.1660 0.1635 0.1611 - 0 . 8 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.1867 - 0 . 7 0.2420 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2177 0.2148 - 0 . 6 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.2546 0.2514 0.2483 0.2451 - 0 . 5 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.2776 - 0 . 4 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3228 0.3192 0.3156 0.3121 - 0 . 3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3557 0.3520 0.3483 - 0 . 2 0.4207 0.4168 0.4129 0.4090 0.4052 0.4013 0.3974 0.3936 0.3897 0.3859 - 0 . 1 0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4325 0.4286 0.4247 - 0 . 0 0.5000 0.4960 0.4920 0.4880 0.4840 0.4801 0.4761 0.4721 0.4681 0.4641 0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141 0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517 0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879 0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224 0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549 0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852 0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133 0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389 1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621 1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830 1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015 1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633 1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706 1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767 2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817 2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857 2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890 2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916 2.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936 2.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952 2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.9964 2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.9974 2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.9981 2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986 1

t distribution critical values - t ⇤ 0 t ⇤ p = P ( t ⌫ ≥ t ⇤ ) Key: Table entry for p and C is the critical value t ⇤ with probability p lying to its right and probability C lying between - t ⇤ and t ⇤ . Upper tail probability p ⌫ .25 .20 .15 .10 .05 .025 .02 .01 .005 .0025 .001 .0005 1 1.000 1.376 1.963 3.078 6.314 12.71 15.89 31.82 63.66 127.3 318.3 636.6 2 0.816 1.061 1.386 1.886 2.920 4.303 4.849 6.965 9.925 14.09 22.33 31.60 3 0.765 0.978 1.250 1.638 2.353 3.182 3.482 4.541 5.841 7.453 10.21 12.92 4 0.741 0.941 1.190 1.533 2.132 2.776 2.999 3.747 4.604 5.598 7.173 8.610 5 0.727 0.920 1.156 1.476 2.015 2.571 2.757 3.365 4.032 4.773 5.893 6.869 6 0.718 0.906 1.134 1.440 1.943 2.447 2.612 3.143 3.707 4.317 5.208 5.959 7 0.711 0.896 1.119 1.415 1.895 2.365 2.517 2.998 3.499 4.029 4.785 5.408 8 0.706 0.889 1.108 1.397 1.860 2.306 2.449 2.896 3.355 3.833 4.501 5.041 9 0.703 0.883 1.100 1.383 1.833 2.262 2.398 2.821 3.250 3.690 4.297 4.781 10 0.700 0.879 1.093 1.372 1.812 2.228 2.359 2.764 3.169 3.581 4.144 4.587 11 0.697 0.876 1.088 1.363 1.796 2.201 2.328 2.718 3.106 3.497 4.025 4.437 12 0.695 0.873 1.083 1.356 1.782 2.179 2.303 2.681 3.055 3.428 3.930 4.318 13 0.694 0.870 1.079 1.350 1.771 2.160 2.282 2.650 3.012 3.372 3.852 4.221 14 0.692 0.868 1.076 1.345 1.761 2.145 2.264 2.624 2.977 3.326 3.787 4.140 15 0.691 0.866 1.074 1.341 1.753 2.131 2.249 2.602 2.947 3.286 3.733 4.073 16 0.690 0.865 1.071 1.337 1.746 2.120 2.235 2.583 2.921 3.252 3.686 4.015 17 0.689 0.863 1.069 1.333 1.740 2.110 2.224 2.567 2.898 3.222 3.646 3.965 18 0.688 0.862 1.067 1.330 1.734 2.101 2.214 2.552 2.878 3.197 3.610 3.922 19 0.688 0.861 1.066 1.328 1.729 2.093 2.205 2.539 2.861 3.174 3.579 3.883 20 0.687 0.860 1.064 1.325 1.725 2.086 2.197 2.528 2.845 3.153 3.552 3.850 21 0.686 0.859 1.063 1.323 1.721 2.080 2.189 2.518 2.831 3.135 3.527 3.819 22 0.686 0.858 1.061 1.321 1.717 2.074 2.183 2.508 2.819 3.119 3.505 3.792 23 0.685 0.858 1.060 1.319 1.714 2.069 2.177 2.500 2.807 3.104 3.485 3.768 24 0.685 0.857 1.059 1.318 1.711 2.064 2.172 2.492 2.797 3.091 3.467 3.745 25 0.684 0.856 1.058 1.316 1.708 2.060 2.167 2.485 2.787 3.078 3.450 3.725 26 0.684 0.856 1.058 1.315 1.706 2.056 2.162 2.479 2.779 3.067 3.435 3.707 27 0.684 0.855 1.057 1.314 1.703 2.052 2.158 2.473 2.771 3.057 3.421 3.690 28 0.683 0.855 1.056 1.313 1.701 2.048 2.154 2.467 2.763 3.047 3.408 3.674 29 0.683 0.854 1.055 1.311 1.699 2.045 2.150 2.462 2.756 3.038 3.396 3.659 30 0.683 0.854 1.055 1.310 1.697 2.042 2.147 2.457 2.750 3.030 3.385 3.646 40 0.681 0.851 1.050 1.303 1.684 2.021 2.123 2.423 2.704 2.971 3.307 3.551 50 0.679 0.849 1.047 1.299 1.676 2.009 2.109 2.403 2.678 2.937 3.261 3.496 60 0.679 0.848 1.045 1.296 1.671 2.000 2.099 2.390 2.660 2.915 3.232 3.460 80 0.678 0.846 1.043 1.292 1.664 1.990 2.088 2.374 2.639 2.887 3.195 3.416 100 0.677 0.845 1.042 1.290 1.660 1.984 2.081 2.364 2.626 2.871 3.174 3.390 1000 0.675 0.842 1.037 1.282 1.646 1.962 2.056 2.330 2.581 2.813 3.098 3.300 .50 .60 .70 0.80 .90 .95 .96 .98 .99 .995 .998 .999 Probability C - t ⇤ 0 t ⇤ C 1