The data was obtained from car crash experiments. The table values are the chest deceleration data (g) for the dummy in the driver’s seat. Use a 0.05 significance level to test the null hypothesis that the different weight categories have the same mean. Do the data suggest that larger cars are safer?
Subcompact: 56,57,56,60,55
Compact: 46,51,54,48,51
Midsize: 51,53,48,45,45
Full-size: 45,48,40,44,51
Should null hypothesis be rejected? (circle)
YES/NO because the P-value is GREATER/LESS than the significance level
Do the data suggest that larger cars are safer? (circle)
YES/NO because the null hypothesis IS/IS NOT rejected.
disregard I figured it out.
To test the null hypothesis that the different weight categories have the same mean, we can perform an analysis of variance (ANOVA) test. Here are the steps to follow:
1. Calculate the mean chest deceleration for each weight category:
Subcompact: (56+57+56+60+55)/5 = 56.8
Compact: (46+51+54+48+51)/5 = 50
Midsize: (51+53+48+45+45)/5 = 48.4
Full-size: (45+48+40+44+51)/5 = 45.6
2. Calculate the overall mean chest deceleration:
(56.8+50+48.4+45.6)/4 ≈ 50.2
3. Calculate the sum of squares within groups (SSW):
SSW = (56-56.8)² + (57-56.8)² + ... + (51-50.2)² + ... + (51-50.2)² + ...
4. Calculate the sum of squares between groups (SSB):
SSB = 5 * ((56.8-50.2)² + (50-50.2)² + (48.4-50.2)² + (45.6-50.2)²)
5. Calculate the degrees of freedom within groups (dofW):
dofW = (number of observations) - (number of groups) = (5 + 5 + 5 + 5) - 4 = 16
6. Calculate the degrees of freedom between groups (dofB):
dofB = (number of groups) - 1 = 4 - 1 = 3
7. Calculate the mean square within groups (MSW):
MSW = SSW / dofW
8. Calculate the mean square between groups (MSB):
MSB = SSB / dofB
9. Calculate the F statistic:
F = MSB / MSW
10. Look up the critical F value for α = 0.05 and (dofB, dofW) in the F-distribution table.
11. Compare the calculated F value with the critical F value:
- If the calculated F value is greater than the critical F value, we reject the null hypothesis.
(Circle YES because the null hypothesis should be rejected.)
- If the calculated F value is less than or equal to the critical F value, we fail to reject the null hypothesis.
(Circle NO because the null hypothesis should not be rejected.)
12. Finally, assess whether larger cars are safer based on the result:
- If the null hypothesis is rejected, it suggests that the mean chest deceleration is significantly different between weight categories. Therefore, it is possible to say that larger cars are safer.
(Circle YES because the null hypothesis is rejected.)
- If the null hypothesis is not rejected, it suggests that there is not enough evidence to conclude that the mean chest deceleration is different between weight categories. Therefore, it is not possible to definitively say that larger cars are safer.
(Circle NO because the null hypothesis is not rejected.)
Please perform the calculations and compare the calculated F value with the critical F value to determine whether the null hypothesis should be rejected and whether larger cars are safer.
To test whether the different weight categories have the same mean and determine if larger cars are safer, we can perform an analysis of variance (ANOVA) test.
To start, calculate the means of each weight category:
Subcompact: (56+57+56+60+55)/5 = 56.8
Compact: (46+51+54+48+51)/5 = 50
Midsize: (51+53+48+45+45)/5 = 48.4
Full-size: (45+48+40+44+51)/5 = 45.6
Next, calculate the grand mean by taking the average of all the data points:
(56+57+56+60+55+46+51+54+48+51+51+53+48+45+45+45+48+40+44+51)/20 = 49.9
Now, calculate the sum of squares within groups (SSW):
For the Subcompact group:
(56-56.8)^2 + (57-56.8)^2 + (56-56.8)^2 + (60-56.8)^2 + (55-56.8)^2 = 6.4
For the Compact group:
(46-50)^2 + (51-50)^2 + (54-50)^2 + (48-50)^2 + (51-50)^2 = 7.2
For the Midsize group:
(51-48.4)^2 + (53-48.4)^2 + (48-48.4)^2 + (45-48.4)^2 + (45-48.4)^2 = 10.4
For the Full-size group:
(45-45.6)^2 + (48-45.6)^2 + (40-45.6)^2 + (44-45.6)^2 + (51-45.6)^2 = 16.4
SSW = 6.4 + 7.2 + 10.4 + 16.4 = 40.4
Now, calculate the sum of squares between groups (SSB):
SSB = (5*(56.8-49.9)^2) + (5*(50-49.9)^2) + (5*(48.4-49.9)^2) + (5*(45.6-49.9)^2)
= 140.2
The degrees of freedom (df) for both SSW and SSB are calculated as follows:
df_within = total number of data points - total number of groups = 20 - 4 = 16
df_between = total number of groups - 1 = 4 - 1 = 3
Now, calculate the mean squares (MS):
MS_within = SSW / df_within = 40.4 / 16 = 2.525
MS_between = SSB / df_between = 140.2 / 3 = 46.73
To determine if the null hypothesis should be rejected, calculate the F-statistic:
F = MS_between / MS_within = 46.73 / 2.525 = 18.49
Next, compute the p-value associated with the F-statistic. In this case, we have a significance level of 0.05. Using statistical software or an F-table, we find that the p-value associated with an F-statistic of 18.49 and 3 and 16 degrees of freedom is significantly less than 0.05 (p < 0.05).
Therefore, we can reject the null hypothesis. The data suggest that the different weight categories do not have the same mean acceleration. However, this analysis does not directly indicate whether larger cars are safer. It only provides evidence that the chest deceleration is different among the weight categories. To determine if larger cars are safer, it would be necessary to analyze other relevant factors such as crash test ratings, safety features, etc.