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question:A manufacturer produces three different brands of batteries. In order to determine if there is a significant difference in the lifespan of the batteries, a random sample of each brand is taken and their lifespans are recorded. The data is shown below: Brand A: 1200, 1400, 1300, 1500, 1250 Brand B: 1150, 1400, 1350, 1200, 1300 Brand C: 1300, 1100, 1350, 1400, 1200 Determine whether there is a significant difference in the lifespan of the three battery brands at a significance level of 0.05 using ANOVA.

answer:To determine if there is a significant difference in the lifespan of the three battery brands, we will perform a one-way ANOVA test. Step 1: Calculate the means of each group and the overall mean. Brand A mean (M_A) = (1200 + 1400 + 1300 + 1500 + 1250) / 5 = 6650 / 5 = 1330 Brand B mean (M_B) = (1150 + 1400 + 1350 + 1200 + 1300) / 5 = 6400 / 5 = 1280 Brand C mean (M_C) = (1300 + 1100 + 1350 + 1400 + 1200) / 5 = 6350 / 5 = 1270 Overall mean (M_T) = (6650 + 6400 + 6350) / 15 = 19400 / 15 = 1293.33 Step 2: Calculate the sum of squares between groups (SSB), sum of squares within groups (SSW), and total sum of squares (SST). SSB = Σn_i(M_i - M_T)^2 SSB = 5(1330 - 1293.33)^2 + 5(1280 - 1293.33)^2 + 5(1270 - 1293.33)^2 SSB = 5(36.67^2) + 5(-13.33^2) + 5(-23.33^2) = 5(1344.44) + 5(177.77) + 5(543.44) = 6722.22 + 888.85 + 2717.22 = 10328.29 SSW = ΣΣ(x_ij - M_i)^2 SSW = (1200-1330)^2 + (1400-1330)^2 + ... + (1200-1270)^2 SSW = 16900 + 4900 + 900 + 28900 + 6400 + 16900 + 4900 + 900 + 16900 + 4900 SSW = 16900 + 4900 + 900 + 28900 + 6400 + 16900 + 4900 + 900 + 16900 + 4900 = 68400 SST = SSB + SSW = 10328.29 + 68400 = 78728.29 Step 3: Calculate the degrees of freedom for each sum of squares. df_SSB = k - 1 = 3 - 1 = 2 df_SSW = N - k = 15 - 3 = 12 df_SST = N - 1 = 15 - 1 = 14 Step 4: Calculate the mean squares for each sum of squares. MSB = SSB / df_SSB = 10328.29 / 2 = 5164.14 MSW = SSW / df_SSW = 68400 / 12 = 5700 Step 5: Calculate the F-statistic. F = MSB / MSW = 5164.14 / 5700 = 0.906 Step 6: Determine the critical F-value at a significance level of 0.05. Using an F-distribution table, we find the critical F-value for df_SSB = 2 and df_SSW = 12 at a significance level of 0.05 is approximately 3.89. Step 7: Compare the F-statistic to the critical F-value. Since the F-statistic (0.906) is less than the critical F-value (3.89), we fail to reject the null hypothesis. Conclusion: At a significance level of 0.05, there is no significant difference in the lifespan of the three battery brands.

question:A company wants to test the effectiveness of three different training programs for their employees. They randomly select 15 employees and assign 5 to each training program. After completing the programs, the employees are given a test and their scores are recorded. The ANOVA table shows the sum of squares values for between groups and within groups as 100 and 225, respectively. Find the F-statistic and conclude whether there is a significant difference in the effectiveness of the three training programs.

answer:To find the F-statistic, we first need to calculate the mean square values for between groups (MSB) and within groups (MSW). Total number of groups (k) = 3 Total number of employees (n) = 15 Number of employees per group (n_i) = 5 Degrees of freedom for between groups (dfB) = k - 1 = 3 - 1 = 2 Degrees of freedom for within groups (dfW) = n - k = 15 - 3 = 12 Now, we can calculate the mean square values: MSB (Mean Square Between) = Sum of Squares Between (SSB) / dfB = 100 / 2 = 50 MSW (Mean Square Within) = Sum of Squares Within (SSW) / dfW = 225 / 12 = 18.75 Now, we can calculate the F-statistic: F-statistic = MSB / MSW = 50 / 18.75 = 2.67 To determine if there is a significant difference in the effectiveness of the three training programs, we need to compare the F-statistic to the critical F-value at a specific significance level (usually 0.05). We can use an F-distribution table to find the critical F-value for dfB = 2 and dfW = 12 at a 0.05 significance level. From the F-distribution table, the critical F-value is approximately 3.89. Since our calculated F-statistic (2.67) is less than the critical F-value (3.89), we fail to reject the null hypothesis. This means that there is no significant difference in the effectiveness of the three training programs at the 0.05 significance level.

question:magine a study to analyze the effect of different types of fertilizer on plant growth. 4 groups of 10 identical plants were used for the study. Group 1 was treated with Fertilizer A, Group 2 was treated with Fertilizer B, Group 3 was treated with Fertilizer C, and Group 4 was left untreated as a control group. After 6 weeks, the heights of the plants were measured and recorded. The results are as follows: Group 1 (Fertilizer A): mean = 11.2 inches, variance = 0.98 Group 2 (Fertilizer B): mean = 12.3 inches, variance = 1.03 Group 3 (Fertilizer C): mean = 10.5 inches, variance = 0.95 Group 4 (Control): mean = 8.6 inches, variance = 0.99 Using ANOVA and a significance level of 0.05, can we conclude that there is a significant difference in plant growth between the groups treated with different fertilizers and the control group?

answer:To determine if there is a significant difference in plant growth between the groups treated with different fertilizers and the control group, we will perform a one-way ANOVA test. Step 1: Calculate the overall mean (Grand Mean) The overall mean is the average of all the individual group means. Grand Mean = (11.2 + 12.3 + 10.5 + 8.6) / 4 = 42.6 / 4 = 10.65 inches Step 2: Calculate the Sum of Squares Between Groups (SSB) and Sum of Squares Within Groups (SSW) SSB = Σn_i (mean_i - Grand Mean)^2 SSW = Σ(n_i - 1) * variance_i For Group 1 (Fertilizer A): SSB_1 = 10 * (11.2 - 10.65)^2 = 10 * 0.55^2 = 3.025 SSW_1 = (10 - 1) * 0.98 = 9 * 0.98 = 8.82 For Group 2 (Fertilizer B): SSB_2 = 10 * (12.3 - 10.65)^2 = 10 * 1.65^2 = 27.225 SSW_2 = (10 - 1) * 1.03 = 9 * 1.03 = 9.27 For Group 3 (Fertilizer C): SSB_3 = 10 * (10.5 - 10.65)^2 = 10 * (-0.15)^2 = 0.225 SSW_3 = (10 - 1) * 0.95 = 9 * 0.95 = 8.55 For Group 4 (Control): SSB_4 = 10 * (8.6 - 10.65)^2 = 10 * (-2.05)^2 = 42.025 SSW_4 = (10 - 1) * 0.99 = 9 * 0.99 = 8.91 Total SSB = SSB_1 + SSB_2 + SSB_3 + SSB_4 = 3.025 + 27.225 + 0.225 + 42.025 = 72.5 Total SSW = SSW_1 + SSW_2 + SSW_3 + SSW_4 = 8.82 + 9.27 + 8.55 + 8.91 = 35.55 Step 3: Calculate the Mean Squares Between Groups (MSB) and Mean Squares Within Groups (MSW) MSB = SSB / (number of groups - 1) = 72.5 / (4 - 1) = 72.5 / 3 = 24.167 MSW = SSW / (total number of samples - number of groups) = 35.55 / (40 - 4) = 35.55 / 36 = 0.987 Step 4: Calculate the F-statistic F = MSB / MSW = 24.167 / 0.987 = 24.48 Step 5: Determine the critical F-value Using an F-distribution table with a significance level of 0.05, degrees of freedom between groups (df1) = 3, and degrees of freedom within groups (df2) = 36, we find the critical F-value to be approximately 2.87. Step 6: Compare the F-statistic to the critical F-value Since the F-statistic (24.48) is greater than the critical F-value (2.87), we reject the null hypothesis that there is no significant difference in plant growth between the groups treated with different fertilizers and the control group. Therefore, we can conclude that there is a significant difference in plant growth between the groups treated with different fertilizers and the control group at a significance level of 0.05.

question:A car manufacturing company is testing the fuel efficiency of three different models of cars. They tested 8 cars of each model and recorded the distance traveled in miles with 10 gallons of fuel. The data is given below. Model 1: 240, 220, 260, 250, 270, 230, 240, 250 Model 2: 280, 290, 300, 270, 260, 280, 295, 285 Model 3: 310, 340, 320, 330, 350, 320, 340, 350 Is there a significant difference in the fuel efficiency of the three models of cars at a 5% level of significance using ANOVA?

answer:To determine if there is a significant difference in the fuel efficiency of the three models of cars at a 5% level of significance using ANOVA, we need to perform the following steps: 1. Calculate the mean of each group (model). 2. Calculate the overall mean. 3. Calculate the sum of squares between groups (SSB), sum of squares within groups (SSW), and total sum of squares (SST). 4. Calculate the mean square between groups (MSB) and mean square within groups (MSW). 5. Calculate the F statistic and compare it to the critical F value. Let's perform these calculations: 1. Mean of each group: Model 1: (240+220+260+250+270+230+240+250)/8 = 246.25 Model 2: (280+290+300+270+260+280+295+285)/8 = 281.25 Model 3: (310+340+320+330+350+320+340+350)/8 = 332.5 2. Overall mean: (246.25+281.25+332.5)/3 = 286.67 3. Calculate SSB, SSW, and SST: SSB = Σn_i*(mean_i - overall_mean)^2 SSB = 8*((246.25-286.67)^2 + (281.25-286.67)^2 + (332.5-286.67)^2) = 8*(1640.56 + 29.16 + 2102.56) = 21872.8 SSW = ΣΣ(x_ij - mean_i)^2 SSW = (20^2 + 30^2 + 10^2 + 0^2 + 20^2 + 10^2 + 0^2 + 0^2) + (10^2 + 0^2 + 20^2 + 10^2 + 20^2 + 0^2 + 10^2 + 0^2) + (20^2 + 0^2 + 10^2 + 20^2 + 0^2 + 10^2 + 0^2 + 0^2) = 3700 SST = SSB + SSW = 21872.8 + 3700 = 25572.8 4. Calculate MSB and MSW: MSB = SSB/(k-1) = 21872.8/(3-1) = 10936.4 MSW = SSW/(N-k) = 3700/(24-3) = 185 5. Calculate the F statistic and compare it to the critical F value: F = MSB/MSW = 10936.4/185 = 59.11 Now, we need to find the critical F value for a 5% level of significance. We use the F-distribution table with degrees of freedom (df1 = k-1 = 2, df2 = N-k = 21). The critical F value for a 5% level of significance is 3.47. Since the calculated F statistic (59.11) is greater than the critical F value (3.47), we reject the null hypothesis that there is no significant difference in the fuel efficiency of the three models of cars. Therefore, there is a significant difference in the fuel efficiency of the three models of cars at a 5% level of significance using ANOVA.