A random sample of 160 car crashes are selected and categorized by age. The results are listed below. The age distribution of dr
1 answer:
Answer:
The claim that all ages have purchase rates proportional to their driving rates is false.
Step-by-step explanation:
The complete question is:
A random sample of 160 car accidents are selected and categorized by the age of the driver determined to be at fault. 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. Calculate the chi-square test statistic used to test the claim that all ages have crash rates proportional to their driving rates.
Age >26 26-45 46-65 45<
Drivers 66 39 25 30
A) 95.431
B)101.324
C)85.123
D)75.101
Solution:
In this case we need to test whether there is sufficient evidence to warrant rejection of the claim that all ages have crash rates proportional to their driving rates.
A Chi-square test for goodness of fit will be used in this case.
The hypothesis can be defined as:
<em>H₀</em>: The observed frequencies are same as the expected frequencies.
<em>Hₐ</em>: The observed frequencies are not same as the expected frequencies.
The test statistic is given as follows:
The values are computed in the table.
The test statistic value is 75.10.
The degrees of freedom of the test is:
n - 1 = 4 - 1 = 3
Compute the p-value of the test as follows:
p-value < 0.00001*
Use a Chi-square table.
p-value < 0.00001 < α = 0.05.
So, the null hypothesis will be rejected at any significance level.
Thus, there is sufficient evidence to warrant rejection of the claim that ages have crash rates proportional to their driving rates.
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<h2>The required "option B) 
" is correct.</h2>
Step-by-step explanation:
You can refer figure:
To find, the value of x = ?
We know that,
Consecutive angles are supplementary
The diagonals bisect the angles.
Divided by 2 both sides, we get
⇒
⇒
⇒
⇒ 4x =
⇒
⇒
∴ The value of x =
Thus, the required "option B) " is correct.
Answer:
a. 0.4129 = 41.29% of the bank's Visa cardholders pay more than 29 dollars in interest.
b. 0.1867 = 18.67% of the bank's Visa cardholders pay more than 35 dollars in interest.
c. 0.0742 = 7.42% of the bank's Visa cardholders pay less than 14 dollars in interest.
d. An interest payment of $35.2 is exceeded by only 18% of the bank's Visa cardholders.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean of 27 dollars and a standard deviation of 9 dollars.
This means that
A. What proportion of the bank's Visa cardholders pay more than 29 dollars in interest?
This is 1 subtracted by the p-value of Z when X = 29, so:
has a p-value of 0.5871.
1 - 0.5871 = 0.4129
0.4129 = 41.29% of the bank's Visa cardholders pay more than 29 dollars in interest.
B. What proportion of the bank's Visa cardholders pay more than 35 dollars in interest?
This is 1 subtracted by the p-value of Z when X = 35, so:
has a p-value of 0.8133.
1 - 0.8133 = 0.1867
0.1867 = 18.67% of the bank's Visa cardholders pay more than 35 dollars in interest.
C. What proportion of the bank's Visa cardholders pay less than 14 dollars in interest?
This is the p-value of Z when X = 14. So
has a p-value of 0.0742.
0.0742 = 7.42% of the bank's Visa cardholders pay less than 14 dollars in interest.
D. What interest payment is exceeded by only 18% of the bank's Visa cardholders?
This is the 100 - 18 = 82nd percentile, which is X when Z has a p-value of 0.82, so X when Z = 0.915.
An interest payment of $35.2 is exceeded by only 18% of the bank's Visa cardholders.