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.
Answer:
StartFraction 7 over 2 EndFraction can be used to find the unit rate if one divides 7 by 2 and compares the result to 1
Step-by-step explanation:
we know that
The formula to calculate the slope between two points is equal to
we have the points
B(2, 7) and D(4, 14)
substitute the values
The unit rate is
therefore
StartFraction 7 over 2 EndFraction can be used to find the unit rate if one divides 7 by 2 and compares the result to 1
Answer:
a) n(none) = 25
b) n(PE but not Bio) = 25
c) n(ENG but not both BIO and PE) = 55
d) n(students that did not take Eng or Bio) = 40
e) P( Students did not take exactly two subjects) = 0.65
Step-by-step explanation:
From the Venn diagram drawn:
a) Number of students that took none
n(Freshmen) = 135
n(all three) = 5
n (PE and Bio) = 10
n(PE and Eng) = 15
n(Bio and Eng) = 7
n (PE and Bio only) = 10 - 5 = 5
n(PE and Eng only) = 15 - 5 = 10
n(Bio and Eng only) = 7 - 5 = 2
n(PE only) = 35 - 5 - 5 - 10 = 15
n(Bio only) = 42 - 5 - 5 - 2 = 30
n(Eng only) = 60 - 10 - 5 -2 = 43
n(Freshmen) = n(PE only) + n(Bio only) + n(Eng only) + n(PE and Bio only) + n(PE and Eng only) + n(Bio and Eng only) + n(all three) + n(none)
135 = 15 + 30 + 43 + 5 + 10 + 2 + 5 + n(none)
135 = 110 + n(none)
n(none) = 135 - 110
n(none) = 25
b)Number of students that too PE but not Bio
n(PE but not bio)= n(PE only) + n(PE and Eng only)
n(PE but not Bio) = 15 + 10
n(PE but not Bio) = 25
c) Number of students that took ENG but not both BIO and PE
n(ENG but not both BIO and PE) = n(Eng only) + n(Eng and Bio only) + n(Eng and PE only) = 43 + 2 + 10
n(ENG but not both BIO and PE) = 55
d) Number of students that did not take ENG or BIO
n( students that did not take Eng or Bio) = n(PE only) + n(none)
n(students that did not take Eng or Bio) = 15 + 25
n(students that did not take Eng or Bio) = 40
e) Probability that a randomly-chosen student from this group did not take exactly two subjects
n( Students that did not take exactly two subjects) = n(PE only) + n(Bio only) + n(Eng only)
n( Students that did not take exactly two subjects) = 15 + 30 + 43
n( Students that did not take exactly two subjects) = 88
P( Students did not take exactly two subjects) = 88/135
P( Students did not take exactly two subjects) = 0.65
