Sangram Keshari Patro

Business Report - 2 PG Program in Data Science and Business Analytics submitted by Sangram Keshari Patro BATCH:PGPDSBA.O.AUG24.B Contents 1 Problem 1 1.1 What is the probability that a randomly chosen player would suer an injury? . . . . . . 1.2 What is the probability that a player is a forward or a winger? . . . . . . . . . . . . . . 1.3 What is the probability that a randomly chosen player plays in a striker position and has foot injury? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 What is the probability that a randomly chosen injured player is a striker? . . . . . . . . . . a . . . . . . 2 Problem 2 2.1 What proportion of the gunny bags have a breaking strength of less than 3.17 kg per sq cm? 2.2 What proportion of the gunny bags have a breaking strength of at least 3.6 kg per sq cm.? . 2.3 What proportion of the gunny bags have a breaking strength between 5 and 5.5 kg per sq cm.? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 What proportion of the gunny bags have a breaking strength NOT between 3 and 7.5 kg per sq cm.? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Problem 3 3.1 Zingaro has reason to believe that the unpolished stones may not be suitable for printing. Do you think Zingaro is justied in thinking so? . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Is the mean hardness of the polished and unpolished stones the same? . . . . . . . . . . . . 4 Problem 4 4.1 How does the hardness of implants vary depending on dentists? . . . . . . . . . . . . . . 4.1.1 Null Hypothesis (H0 ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.2 Alternate Hypothesis (HA ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.3 Assumptions of the One-Way ANOVA . . . . . . . . . . . . . . . . . . . . . . . . 4.1.4 Analysis and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 How does the hardness of implants vary depending on methods? . . . . . . . . . . . . . . 4.2.1 Assumptions of the One-Way ANOVA . . . . . . . . . . . . . . . . . . . . . . . . 4.2.2 Tukey HSD Test for Alloy 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.3 ANOVA Results for Alloy 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.4 Tukey HSD Test for Alloy 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 What is the interaction eect between the dentist and method on the hardness of dental implants for each type of alloy? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.1 Assumptions of the Two-Way ANOVA . . . . . . . . . . . . . . . . . . . . . . . . 4.3.2 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 How does the hardness of implants vary depending on dentists and methods together? . . 4.4.1 ANOVA Results for Alloy 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.2 Tukey HSD Test for Alloy 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.3 ANOVA Results for Alloy 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.4 Tukey HSD Test for Alloy 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 4 4 4 4 4 5 5 5 6 6 7 7 7 7 . . . . . . . . . . . . - . . . . . . . . - List of Figures- Proportion of Gunny Bags with Breaking Strength Less than 3.17 kg/sq cm . . . . . . . . Proportion of Gunny Bags with Breaking Strength of at least 3.6 kg per sq cm. . . . . . . Proportion of Gunny Bags with Breaking Strength between 5 and 5.5 kg per sq cm. . . . Proportion of Gunny Bags with Breaking Strength NOT between 3 and 7.5 kg per sq cm. Hardness of implants vary depending on dentists for two dierent alloys . . . . . . . . . . Hardness of implants vary depending on methods for two dierent alloys . . . . . . . . . . Interaction Plots of how Hardness of implants vary depending on methods & dentists for two dierent alloys . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 . . . . . . 5 5 6 6 8 9 . 11 List of Tables- Data on Players and Injuries . . . . . . . . . . . . ANOVA Results for Alloy-1 . . . . . . . . . . . . . ANOVA Results for Alloy-2 . . . . . . . . . . . . . ANOVA Results for Alloy-1 . . . . . . . . . . . . . Tukey HSD Test Results for Alloy 1 . . . . . . . . ANOVA Results for Alloy 2 . . . . . . . . . . . . . Tukey HSD Test Results for Alloy 2 . . . . . . . . ANOVA Results for Interaction Eect . . . . . . . ANOVA Table . . . . . . . . . . . . . . . . . . . . ANOVA Results for Dentist and Method for alloy 1 ANOVA Results for Dentist and Method for alloy 2 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . - 1 Problem 1 A physiotherapist with a male football team is interested in studying the relationship between foot injuries and the positions at which the players play from the data collected: Position Striker Forward Attacking Midelder Winger Players Injured Players Not Injured Total 145 90 235 - Total - - Table 1: Data on Players and Injuries 1.1 What is the probability that a randomly chosen player would suer an injury? The probability that a randomly chosen player would suer an injury is calculated as follows: P (Injured) = 145 Total Injured Players = Total Players 235 Simplifying this gives: P (Injured) = 0.617 Thus, the probability that a randomly chosen player would suer an injury is 0.617 or 61.7%. 1.2 What is the probability that a player is a forward or a winger? The probability that a randomly chosen player is either a forward or a winger is: P (Forward or Winger) = Total Forwards + Total Wingers 94 + 29 = Total Players 235 Simplifying this gives: P (Forward or Winger) = 123 =- Thus, the probability that a randomly chosen player is a forward or a winger is 0.523 or 52.3%. 1.3 What is the probability that a randomly chosen player plays in a striker position and has a foot injury? The probability that a randomly chosen player plays in a striker position and has a foot injury is: P (Striker and Injured) = 45 Injured Strikers = Total Players 235 Simplifying this gives: P (Striker and Injured) = 0.191 Thus, the probability that a randomly chosen player plays in a striker position and has a foot injury is 0.191 or 19.1%. 1.4 What is the probability that a randomly chosen injured player is a striker? The probability that a randomly chosen injured player is a striker is: P (Striker | Injured) = Injured Strikers 45 = Total Injured Players 145 Simplifying this gives: P (Striker | Injured) = 0.310 Thus, the probability that a randomly chosen injured player is a striker is 0.310 or 31.0%. 4 2 Problem 2 2.1 What proportion of the gunny bags have a breaking strength of less than 3.17 kg per sq cm? Figure 1: Proportion of Gunny Bags with Breaking Strength Less than 3.17 kg/sq cm Here is the visualization that shows the normal distribution of breaking strength for the gunny bags. The shaded red area represents the proportion of gunny bags with a breaking strength of less than 3.17 kg per sq cm, which is approximately 11.12%. 2.2 What proportion of the gunny bags have a breaking strength of at least 3.6 kg per sq cm.? Figure 2: Proportion of Gunny Bags with Breaking Strength of at least 3.6 kg per sq cm. 5 Here is the visualization that shows the normal distribution of breaking strength for the gunny bags. The shaded green region shows the proportion of bags whose strength is 3.6 kg/sq cm or more, representing about 82.47%. 2.3 What proportion of the gunny bags have a breaking strength between 5 and 5.5 kg per sq cm.? Figure 3: Proportion of Gunny Bags with Breaking Strength between 5 and 5.5 kg per sq cm. Here is the visualization that shows the normal distribution of breaking strength for the gunny bags. The shaded orange region shows the proportion of bags within this range, approximately 13.06%. 2.4 What proportion of the gunny bags have a breaking strength NOT between 3 and 7.5 kg per sq cm.? Figure 4: Proportion of Gunny Bags with Breaking Strength NOT between 3 and 7.5 kg per sq cm. 6 Here is the visualization that shows the normal distribution of breaking strength for the gunny bags. The shaded purple regions show the proportion of bags whose strength falls outside the 3 to 7.5 kg/sq cm range, about 13.9%. 3 Problem 3 3.1 Zingaro has reason to believe that the unpolished stones may not be suitable for printing. Do you think Zingaro is justied in thinking so? Null Hypothesis (H0 ): The mean Brinell hardness index of unpolished stones is greater than or equal to 150 (i.e., the stones are suitable for printing). H0 : µ ≥ 150 (Unpolished stones are suitable for printing) Alternative Hypothesis (HA ): The mean Brinell hardness index of unpolished stones is less than 150 (i.e., the stones may not be suitable for printing). HA : µ < 150 (Unpolished stones are not suitable for printing) Signicance Level: We set the signicance level at α = 0.05. Conducting the t-test: We use a one-sample t-test to compare the sample mean hardness of unpolished stones to the population mean of 150. The t-test provides a t-statistic and p-value to be -4.16 and 4.17e-05 respectively. Decision Rule: - The p-value is less than α = 0.05 hence we reject the null hypothesis i.e., the stones may not be suitable for printing 3.2 Is the mean hardness of the polished and unpolished stones the same? Hypotheses: Null Hypothesis (H0 ): The mean Brinell hardness index of polished stones is equal to the mean Brinell hardness index of unpolished stones. H0 : µpolished = µunpolished Alternative Hypothesis (HA ): The mean Brinell hardness index of polished stones is dierent from the mean Brinell hardness index of unpolished stones. HA : µpolished 6= µunpolished Signicance Level: We set the signicance level at α = 0.05. Test Statistic: We use a two-sample t-test to compare the means of the two groups. The t-statistic is calculated as: x̄polished − x̄unpolished t= r s2polished npolished + s2unpolished nunpolished Decision Rule: - The t-test provides a t-statistic and p-value to be -3.24 and 0.0016 respectively. The p-value is less than α = 0.05 hence we reject the null hypothesis i.e.,The mean Brinell hardness index of polished stones is dierent from the mean Brinell hardness index of unpolished stones. 4 Problem 4 4.1 How does the hardness of implants vary depending on dentists? In this analysis, we aim to determine how the hardness of dental implants varies depending on the dentists. The response variable of interest is the hardness of dental implants, and we will analyze this separately for Alloy 1 and Alloy 2. Signicance Level: We set the signicance level at α = 0.05. For both types of alloys, we consider the following hypotheses: 7 4.1.1 Null Hypothesis (H0 ) The hardness of implants does not dier between dentists: H0 : µ1 = µ2 = µ3 = µ4 = µ5 4.1.2 Alternate Hypothesis (HA ) The hardness of implants diers between at least one pair of dentists: HA : At least one µi diers from the others. Figure 5: Hardness of implants vary depending on dentists for two dierent alloys 4.1.3 Assumptions of the One-Way ANOVA Before conducting the ANOVA, the following assumptions must be checked: ˆ Normality: The response variable (hardness) is normally distributed for each group (dentist). ˆ Homogeneity of variance: The variances of hardness between dentists are equal. ˆ Independence: Observations are independent. Note : The p-values of Shapiro's test indicate the distribution is normal for both alloys (except one case for Alloy 2 with Dentist 4) and the population variances are equal for both alloys between dentists. Even if the assumptions are not fully met, we will still proceed with the test as instructed. 4.1.4 Analysis and Results The analysis is conducted separately for Alloy 1 and Alloy 2. ˆ Alloy 1: The results of the ANOVA test indicate whether we can reject the null hypothesis. The p-value is greater than 0.05, hence we accept the null hypothesis and conclude that the implant hardness doesn't dier between dentists for alloy 1. Source C(Dentist) Residual df 4.0 40.0 sum_sq - mean_sq - Table 2: ANOVA Results for Alloy-1 It doesn't dier so we don't require to perform Tukey HSD Test. 8 F - NaN PR(>F) - NaN ˆ Alloy 2: The results of the ANOVA test indicate whether we can reject the null hypothesis. The p-value is greater than 0.05, hence we accept the null hypothesis and conclude that the implant hardness doesn't diers between dentists for alloy 2. Source C(Dentist) Residual df 4.0 40.0 sum_sq -e-e+06 mean_sq - F - NaN PR(>F) - NaN Table 3: ANOVA Results for Alloy-2 It doesn't dier so we don't require to perform Tukey HSD Test. 4.1.5 Conclusion The analysis separately for both Alloy 1 and Alloy 2 provides insights into whether the hardness of dental implants varies based on the dentist. The results of the ANOVA test indicate whether there are no signicant dierences in hardness across dentists, and the same is conrmed by Tukey HSD test. 4.2 How does the hardness of implants vary depending on methods? In this analysis, we aim to determine how the hardness of dental implants varies depending on the methods used. The variable of interest is the hardness of dental implants, and we will analyze this separately for Alloy 1 and Alloy 2. Null Hypothesis (H0 ) The hardness of implants does not dier between methods: H0 : µ1 = µ2 = µ3 Alternate Hypothesis (HA ) The hardness of implants diers between at least one pair of methods: HA : At least one µi dier from the others. Figure 6: Hardness of implants vary depending on methods for two dierent alloys 9 4.2.1 Assumptions of the One-Way ANOVA Before conducting the ANOVA, the following assumptions must be checked: ˆ Normality: The response variable (hardness) is normally distributed for each group (method). ˆ Homogeneity of variance: The variances of hardness between methods are equal. ˆ Independence: Observations are independent. Note : The p-values of Shapiro's test indicate the distribution is normal for both alloys (except one case for Alloy 2 with Method 2) and the population variances are not equal for both alloys between methods. Even if the assumptions are not fully met, we will still proceed with the test as instructed. The analysis is conducted separately for Alloy 1 and Alloy 2. ˆ Alloy 1: The results of the ANOVA test for Alloy 1 are shown in Table 4. Since the p-value is less than 0.05, we reject the null hypothesis for Alloy 1. This indicates that the hardness of implants diers between methods. Source C(Method) Residual df sum_sq 2.0 42.0 - mean_sq - F - NaN PR(>F) - NaN Table 4: ANOVA Results for Alloy-1 4.2.2 Tukey HSD Test for Alloy 1 The Tukey HSD test helps identify which pairs of methods dier signicantly in terms of hardness. For Alloy 1, signicant dierences in hardness are found between Method 1 and Method 3, as well as Group 1 Group 2 Mean Dierence 1 1 2 2 3 3 -6.1333 - - p-adj - Lower Bound Upper Bound Reject -102.714 - - 90.4473 -28.2193 -22.0860 False True True Table 5: Tukey HSD Test Results for Alloy 1 between Method 2 and Method 3. However, the dierence between Method 1 and Method 2 is not statistically signicant. ˆ Alloy 2: 4.2.3 ANOVA Results for Alloy 2 The results of the ANOVA test for Alloy 2 are as follows: Source C(Method) Residual df 2.0 42.0 sum_sq - mean_sq - F 16.4108 NaN PR(>F) - NaN Table 6: ANOVA Results for Alloy 2 Since the p-value is much less than 0.05, we reject the null hypothesis for Alloy 2. This indicates that the hardness of implants diers signicantly between methods. 10 Group 1 Group 2 Mean Dierence 1 1 2 2 3 3 27.0000 - - p-adj - Lower Bound Upper Bound Reject -82.4546 - - - -99.3454 - False True True Table 7: Tukey HSD Test Results for Alloy 2 4.2.4 Tukey HSD Test for Alloy 2 The Tukey HSD test helps identify which pairs of methods dier signicantly in terms of hardness. For Alloy 2, signicant dierences in hardness are found between Method 1 and Method 3, as well as between Method 2 and Method 3. However, the dierence between Method 1 and Method 2 is not statistically signicant. 4.2.5 Conclusion The analysis separately for both Alloy 1 and Alloy 2 provides insights into whether the hardness of dental implants varies based on the methods. For **Alloy 1**, the ANOVA test reveals signicant dierences in hardness across the methods (p = 0.004). This indicates that at least one method diers from the others in terms of hardness. The Tukey HSD test conrms that Methods 1 and 3, as well as Methods 2 and 3, have statistically signicant dierences in hardness, while no signicant dierence is found between Methods 1 and 2. For **Alloy 2**, the ANOVA test also shows signicant dierences in hardness across the methods (p =-). Similarly, the Tukey HSD test indicates signicant dierences between Methods 1 and 3, and Methods 2 and 3, while no signicant dierence is observed between Methods 1 and 2. In both alloys, the ANOVA tests highlight that hardness varies signicantly among the methods, and the Tukey HSD test pinpoints the specic pairs of methods where these dierences occur. 4.3 What is the interaction eect between the dentist and method on the hardness of dental implants for each type of alloy? (a) Alloy 1 (b) Alloy 2 Figure 7: Interaction Plots of how Hardness of implants vary depending on methods & dentists for two dierent alloys Alloy 1: ˆ Method 1: Consistent hardness (∼ 800) across dentists, slight drop for dentists 3 and 5. ˆ Method 2: Starts low (∼ 700), slight decline across dentists, with some variation. ˆ Method 3: Sharp drop in hardness for dentist 4, but large variability. Alloy 2: ˆ Method 1: Hardness remains stable (∼ 800) across all dentists. 11 ˆ Method 2: Starts high (∼ 950), gradually decreases as dentists change, showing large variation. ˆ Method 3: Lowest hardness (∼ 700 to ∼ 600), declines steadily across dentists, with large error bars. In this analysis, we aim to determine how the hardness of dental implants is aected by the interaction between the dentist and method. The response variable is the hardness of dental implants, and we will analyze this separately for Alloy 1 and Alloy 2. Null Hypothesis (H0 ) There is no interaction eect between dentist and method on the hardness of dental implants: H0 : There is no signicant interaction between dentist and method. Alternate Hypothesis (HA ) There is an interaction eect between dentist and method on the hardness of dental implants: HA : There is a signicant interaction between dentist and method. 4.3.1 Assumptions of the Two-Way ANOVA Before conducting the ANOVA, the following assumptions must be checked: ˆ Normality: The response variable (hardness) is normally distributed for each combination of dentist and method. ˆ Homogeneity of variance: The variances of hardness between groups (dentist-method combinations) are equal. ˆ Independence: Observations are independent. Note : We have already veried these conditions separately for both methods and alloys, so there is no need to check them explicitly for their combinations. Regardless of whether the assumptions are fully met, we will proceed with the test as directed. The interaction eects are analyzed separately for Alloy 1 and Alloy 2. Alloy 1 The results of the two-way ANOVA for Alloy 1 are shown in Table 8. Source C(Dentist) C(Method) C(Dentist):C(Method) Residual sum_sq - df - F - NaN PR(>F) - NaN Table 8: ANOVA Results for Interaction Eect Since the p-value for the interaction eect is 0.007, we conclude that there is signicant interaction eect between dentist and method on the hardness of implants for Alloy 1. Alloy 2 For Alloy 2, the results of the two-way ANOVA are shown in Table 9. Since the p-value for the interaction eect is 0.09, we conclude that there is no signicant interaction eect between dentist and method on the hardness of implants for Alloy 2. 12 Table 9: ANOVA Table Source C(Dentist) C(Method) C(Dentist):C(Method) Residual 4.3.2 Sum of Squares (sum_sq - Degrees of Freedom (df - F- NaN P R(> F ) - NaN Conclusion The analysis for both Alloy 1 and Alloy 2 provides insights into whether the interaction between dentist and method aects the hardness of dental implants. ˆ For Alloy 1, **Eect of Dentist**: A p-value of- indicates a statistically signicant dierence in implant hardness among dentists, suggesting the choice of dentist signicantly impacts hardness. **Eect of Method**: A p-value of- shows a signicant eect of the method on implant hardness, with dierent methods leading to notably dierent outcomes. **Interaction Eect**: The p-value for the interaction term C(Dentist) : C(M ethod) is-, indicating a signicant interaction eect on implant hardness, implying that the eect of the method varies by dentist. ˆ For Alloy 2, **Eect of Dentist**: A p-value of 0.3718 suggests no signicant dierence in implant hardness among dentists. **Eect of Method**: The p-value of- indicates a signicant eect of the method on implant hardness, with dierent methods yielding signicantly dierent outcomes. **Interaction Eect**: The interaction term C(Dentist) : C(M ethod) has a p-value of 0.0932, indicating a potential interaction eect; however, it does not reach signicance at the 0.05 level, suggesting no signicant inuence on hardness for Alloy 2. 4.4 How does the hardness of implants vary depending on dentists and methods together? In this analysis, we aim to determine how the hardness of dental implants is aected by taking contributions of dentists and methods together, separately for Alloy 1 and Alloy 2. The response variable is the hardness of dental implants. Null Hypotheses (H0 ): ˆ There is no signicant eect of Dentist on the Response: H0 : µDentist1 = µDentist2 = . . . = µDentistn This means that the mean responses for all dentists are the same. ˆ There is no signicant eect of Method on the Response: H0 : µM ethod1 = µM ethod2 = . . . = µM ethodn This means that the mean responses for all methods are the same. 13 Alternate Hypotheses (HA ): ˆ There is a signicant eect of Dentist on the Response: HA : µDentist1 6= µDentist2 6= . . . 6= µDentistn This means that at least one dentist's mean response diers from the others. ˆ There is a signicant eect of Method on the Response: HA : µM ethod1 6= µM ethod2 6= . . . 6= µM ethodn This means that at least one method's mean response diers from the others. Note : We have already veried these conditions separately for both methods and alloys, so there is no need to check them explicitly for their combinations. Regardless of whether the assumptions are fully met, we will proceed with the test as directed. The analysis is conducted separately for Alloy 1 and Alloy 2. Alloy 1: 4.4.1 ANOVA Results for Alloy 1 The results of the two-way ANOVA test for Alloy 1 are shown in Table 10. ˆ For Method: The p-value is-, which is less than 0.05. Therefore, we reject the null hypothesis and conclude that there is a signicant eect of the method on the response. ˆ For Dentist: The p-value is-, which is slightly greater than 0.05. Therefore, we fail to reject the null hypothesis, suggesting that dentist dierences are not statistically signicant at the 0.05 level. Source C(Dentist) C(Method) Residual sum_sq - df - F - NaN PR(>F) - NaN Table 10: ANOVA Results for Dentist and Method for alloy 1 4.4.2 Tukey HSD Test for Alloy 1 The Tukey HSD test helps identify which combinations of dentists and methods dier signicantly in terms of hardness. Look into the support le for the table. The Tukey HSD test further reveals the signicant pairwise comparisons: For the combinations with reject = True signies that at least one dentist's mean response diers from the others. Alloy 2: 4.4.3 ANOVA Results for Alloy 2 The results of the two-way ANOVA test for Alloy 2 are as follows: Source C(Dentist) C(Method) Residual sum_sq df - - F - NaN PR(>F) - NaN Table 11: ANOVA Results for Dentist and Method for alloy 2 The results of the two-way ANOVA test for Alloy 1 are shown in Table 11. 14 ˆ For Method: The p-value is-, which is less than 0.05. Therefore, we reject the null hypothesis and conclude that there is a signicant eect of the method on the response. ˆ For Dentist: The p-value is-, which is slightly greater than 0.05. Therefore, we fail to reject the null hypothesis, suggesting that dentist dierences are not statistically signicant at the 0.05 level. 4.4.4 Tukey HSD Test for Alloy 2 The Tukey HSD test helps identify which combinations of dentists and methods dier signicantly in terms of hardness. Look into the support le for the table. The Tukey HSD test further reveals the signicant pairwise comparisons: For the combinations with reject = True signies that at least one dentist's mean response diers from the others. 15