Question

= Question Help ustomer wants to know whether the mean chest compression resulting from an institute crash test is the same for family cars, passenger vans, and trucks. The to the right were collected from the institute's study. Vehicle Type Family Cars Passenger Vans Trucks Chest Compression (mm) 30 36 37 43 43 32 26 31 27 33 27 28 37 31 33 41 27 34 onduct a one-way ANOVA test on the data. terpret your results from (a) at the 1% significance level. ecide whether presuming that the assumptions of normal populations and equal population standard deviations are met is reasonable. first, let pc. Hy, and Hy be the population means for family cars, passenger vans and trucks, respectively. What are the correct hypotheses for a one-way ANOVA test? 4. Ho Hot HVEHT Hai Hc = Hv=HT OB. Ho: Pc = Hy =HT Hy: Not all the means are equal. C. Ho: Pc = Hv=HT Hai Hc *Hv *HT OD. Ho: PcHvHT H: All the means are equal. conduct a one-way ANOVA test on the data. What is the F-statistic? (Round to two decimal places as needed.) interpret the results of the one-way ANOVA test, first determine the critical value Fa. (Round to two decimal places as needed.) state the conclusion. the F-statistic vl in the relection region v H. The data I sufficient evidence to conclude that the population means are not all the same to select your answer(s). 16 Lr Ps Lrc ty

Answer 1

Solution: The following table is obtained: Group 1 Group 2 Group 3 30 32 27 36 26 28 37 31 37 43 27 31 43 33 33 41 27 34 Sum = 230 176 190 Average = 38.333 29.333 31.667 8944 5208 6088 St. Dev. = 5.046 3.011 3.777 SS = 127.333333333 33 45.3333333333 33 71.3333333333 33 n = 6 6 6 The total sample size is N = 18. Therefore, the total degrees of freedom are: d ftotal = 18-1= 17 Also, the between-groups degrees of freedom are dfbetween = 3-1 = 2, and the within-groups degrees of freedom are: dfwithin = d ftotal - dfvetween = 17 - 2 = 15 First, we need to compute the total sum of values and the grand mean. The following is obtained ΣΧ; = 230 + 176 + 190 = 596 Also, the sum of squared values is = 8944 +5208 + 6088 = 20240 Based on the above calculations, the total sum of squares is computed as follows 2 Stotal = x - 1 N ΣΧ., = 20240 5962 18 = 505.778

The within sum of squares is computed as shown in the calculation below: S Swithin = $Swithingroups 127.33333333333 + 45.333333333333 + 71.333333333333 =244 The between sum of squares is computed directly as shown in the calculation below: The between sum of squares is computed directly as shown in the calculation below: Now that sum of squares are computed, we can proceed with computing the mean sum of squares: 261.778 2 = 130.889 S Swetween M Sbetween = d fbetween S Swithin M Swithin = dfwithin 244 15 16.267 Now that sum of squares are computed, we can proceed with computing the mean sum of squares: 261.778 = 130.889 2 S Swetween M Sbetween = dfbetween SS within MS within = df within 244 15 = 16.267 Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows: M Steen MS within 130.889 16.267 = 8.046 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: M1 = 2 = 3 Ha: Not all means are equal The above hypotheses will be tested using an F-ratio for a one-way ANOVA. (2) Rejection Region Based on the information provided, the significance level is a = 0.01, and the degrees of freedom are dfi = 2 and 0 f2 = 2, therefore, the rejection region for this F-test is R= {F:F> Fe = 6.359). (3) Test Statistics F= M Sbetween M Swithin 130.889 16.267 8.046 (4) Decision about the null hypothesis Since it is observed that F = 8.046 > F = 6.359, it is then concluded that the null hypothesis is rejected. 0.0042 Using the P-value approach: The p-value is p = 0.0042, and since p 0.01, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that not all 3 population means are equal, at the a = 0.01 significance level