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forsale [732]
3 years ago
9

A random sample of 160 car crashes are selected and categorized by age. The results are listed below. The age distribution of dr

ivers for the given categories is 18% for the under 26 group, 39% for the 26-45 group, 31% for the 45-65 group, and 12% for the group over 65. Calculate the chi-square test statistic
χ to test the claim that all ages have crash rates proportional to their driving rates.
1. Under 26 = 66 drivers
2. 26-45 = 39 drivers
3. 46-65 = 25 drivers
4. Over 65 = 30 drivers
Mathematics
1 answer:
sasho [114]3 years ago
5 0

Answer:

25.05

Step-by-step explanation:

Given that a random sample of 160 car crashes are selected and categorized by age. The results are listed below. The age distribution of drivers for the given categories is 18% for the under 26 group, 39% for the 26-45 group, 31% for the 45-65 group, and 12% for the group over 65.

     

<26 26-45 46-65` >65  

Observed  O 66 39 25 30 160

Expected  E 40 40 40 40 160

     

Chi square      

(O-E)^2/E 16.9 0.025 5.625 2.5 25.05

Thus we get chi square test statistic = 25.05  

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madreJ [45]

Answer:

4. x = 16.8

5. x = 35.98

Step-by-step explanation:

For both triangles that you need to work on, use the trigonometric function which is tan and you are dividing opposite by adjacent for the answers.

For number 4, the equation you are setting up is tan 35 = x/24.  Multiply 24 on both sides to get the answer which is 16.8 for the x side.

For number 5, the equation you are setting up is tan 17 = 11/x.  As a result on your calculator, you should get 35.98 for the x side.

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Step-by-step explanation:

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Use the point-slope formula.

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Question 8 of 10
Andreyy89

Median and IQR are the most appropriate measures of center and spread for this data set.

<h3>Why are Median and IQR the most appropriate?</h3>

Among the 3 central tendencies that includes the mean, median and mode; the median is the better measure because of the followings:

  • Mean is affected by extreme values
  • Mean is not correct if more outliers are present
  • Mean may not represent the nature of the data whether skewed right or left.

Also, the median as the middle entry is not affected by extreme items or outliers, so the median is better than mean,

Furthermore, for the measure of spread, the IQR is better since extreme items will show higher std deviation and also some outliers mislead.

Therefore, Option B is correct.

Read more about measures of center

brainly.com/question/10610456

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