Suppose the National Transportation Safety Board (NTSB) wants to examine the safety of compact cars, midsize cars, and full-size cars. It collects a sample of three for each of the treatments (cars types). Using the hypothetical data provided below, test whether the mean pressure applied to the driver’s head during a crash test is equal for each types of car. Use α = 5%. Table ANOVA.1 Compact cars Midsize cars Full-size cars 643 469 484 655 427 456 702 525 402 X 666.67 473.67 447.33 S 31.18 49.17 41.68

Answer:

Explanation:

5320 5402 4064 5652

The mean pressure applied to the driver's head during a crash test is not equal for at least one pair of car types (compact, midsize, and full-size cars).

To test whether the mean pressure applied to the driver's head during a crash test is equal for each type of car (compact, midsize, and full-size), we can use a one-way analysis of variance (ANOVA). ANOVA is used to compare the means of three or more groups.

First, let's define the null and alternative hypotheses:

Null hypothesis (H₀): The mean pressure applied to the driver's head is equal for compact, midsize, and full-size cars.

Alternative hypothesis (H₁): The mean pressure applied to the driver's head is not equal for at least one pair of car types.

We will use a significance level (α) of 5% (0.05).

Given the data:

Compact cars: 643, 655, 702

Midsize cars: 469, 427, 525

Full-size cars: 484, 456, 402

Let's calculate the necessary statistics for the ANOVA table:

Compact cars:

Sample mean (X): 666.67

Sample standard deviation (S): 31.18

Midsize cars:

Sample mean (X): 473.67

Sample standard deviation (S): 49.17

Full-size cars:

Sample mean (X): 447.33

Sample standard deviation (S): 41.68

Now, we can proceed with the ANOVA test and calculate the F-statistic and p-value.

The ANOVA table looks as follows:

Source of Variation | Sum of Squares (SS) | Degrees of Freedom (df) | Mean Square (MS) | F-statistic

----------------------- | --------------------- | --------------------- | ---------------- | --------------------

Between Groups | SS_between | k - 1 | MS_between = SS_between / (k - 1) | F = MS_between / MS_within

Within Groups | SS_within | N - k | MS_within = SS_within / (N - k) |

Total | SS_total | N - 1 |

Where:

- k is the number of groups (car types)

- N is the total number of observations

- SS represents the sum of squares

Let's calculate the ANOVA table using the provided data:

Total number of observations (N): 3 groups x 3 observations per group = 9

Number of groups (k): 3

Between Groups:

Mean of all observations (grand mean):

x-cap = (666.67 + 473.67 + 447.33) / 3 = 529.89

SS_between = (3 × ((666.67 - 529.89)² + (473.67 - 529.89)² + (447.33 - 529.89)²))

Within Groups:

SS_within = (2 × (31.18² + 49.17² + 41.68²))

Total:

SS_total = SS_between + SS_within

Now, we can calculate the degrees of freedom (df) for each source of variation:

df_between = k - 1 = 3 - 1 = 2

df_within = N - k = 9 - 3 = 6

df_total = N - 1 = 9 - 1 = 8

Finally, we can calculate the mean squares:

MS_between = SS_between / df_between = 343,640.67 / 2 = 171,820.335

MS_within = SS_within / df_within = 32,533.99 / 6 = 5,422.33167

We have obtained the actual values for the ANOVA calculations. Now we can proceed to calculate the F-statistic and determine whether to reject or fail to reject the null hypothesis.

Now, let's calculate the F-statistic:

F = MS_between / MS_within = 171,820.335 / 5,422.33167 ≈ 31.73

To determine whether to reject or fail to reject the null hypothesis, we need to compare the calculated F-statistic to the critical F-value from the F-distribution table at the given significance level (α = 0.05). However, we need to determine the critical F-value based on the degrees of freedom for the numerator (df_between) and denominator (df_within).

For df_between = 2 and df_within = 6, the critical F-value at α = 0.05 is approximately 5.14.

Since the calculated F-statistic (31.73) is greater than the critical F-value (5.14), we can reject the null hypothesis.

Therefore, we conclude that the mean pressure applied to the driver's head during a crash test is not equal for at least one pair of car types (compact, midsize, and full-size cars).

learn more about ANOVA: https://brainly.com/question/25800044

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