Answer:
The answer is participlating.
I apologize if I am incorrect.
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~ Myaka O.
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:
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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).
How did we arrive at this assertion?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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Read the following excerpt from The Age of Innocence, and identify the narrator’s tone regarding Countess Olenska’s behavior.
"To the general relief, the Countess Olenska was not present in her grandmother's drawing-room during the visit of the betrothed couple. Mrs. Mingott said she had gone out; which, on a day of such glaring sunlight, and at the "shopping hour," seemed in itself an indelicate thing for a compromised woman to do."
confusion
relief
annoyance
disapproval
Answer:
the tone is confusion
you can tell because the narrator is quite perplexed by why the Countess was not present in her grandmother's room, and why she had gone out during the sunlight at noon ("shopping hour") , since it is strange for her a married woman like her to do so.
Trong giai đoạn biến động kinh tế tại Việt Nam năm 2008 , trong hoàn cảnh lạm phát tang cao , có ý kiến cho rằng số lượng ngân hàng thương mại tại Việt Nam quá nhiều (39 ngân hàng , tương đương số ngân hàng của Indonesia, nhiều hơn Malaysia, Philipines ), tạo ra lượng tiền bút tệ lớn , góp phần làm tăng tỷ lệ lạm phát . Anh/ Chị hãy nêu nhận xết của mình về ý kiến này
Answer:
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Explanation:
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their IQ score just as much as the Experimental Group. Name 2 possible
explanations for these results.
4) Name 2 ways you could improve on the experiment to determine if eating apples
has any special effect on people?
5) In the above experiment, describe how to conduct it as a double-blind study.
Answer:
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