Question 1

A sociologist believes that the levels of educational attainment of homeless persons are not uniformly
distributed. To test this claim, you randomly survey 100 homeless persons and record the educational
attainment of each. The results are shown in the following table. Find the critical value χ $$\chi _ { 0 } { } ^ { 2 }$$
 to test the
sociologistʹs claim. Use $$\alpha$$ = 0.10. $$\begin{array} { l c } 
\hline \text { Response } & \text { Frequency, f } \
\hline \text { Less than high school } & 38 \
\text { High school graduate/G.E.D. } & 34 \
\text { More than high school } & 28 \
\hline
\end{array}$$

Accepted Answer

4.605

Question 2

Many track runners believe that they have a better chance of winning if they start in the inside lane that is closest to the field. For the data below, the lane closest to the field is Lane 1, the next lane is Lane 2, and so on
Until the outermost lane, Lane 6. The data lists the number of wins for track runners in the different starting
Positions. Calculate the chi-square test statistic χ $$\chi ^ { 2 }$$ to test the claim that the number of wins is uniformly
Distributed across the different starting positions. The results are based on 240 wins. $$\begin{array} { c | c c c c c c } 
\text { Starting Position } & 1 & 2 & 3 & 4 & 5 & 6 \
\hline \text { Number of Wins } & 45 & 36 & 33 & 50 & 32 & 44
\end{array}$$
A)6.750
B)9.326
C)12.592
D)15.541

Accepted Answer

A

Question 3

The frequency distribution shows the ages for a sample of 100 employees. Find the expected frequencies for
each class to determine if the employee ages are normally distributed. $$\begin{array} { l c } 
\hline \text { Class boundaries } & \text { Frequency, f } \
\hline 29.5 - 39.5 & 14 \
39.5 - 49.5 & 29 \
49.5 - 59.5 & 31 \
59.5 - 69.5 & 18 \
69.5 - 79.5 & 8 \
\hline
\end{array}$$

Accepted Answer

11, 26, 32, 21, and 7, respectively.

Question 4

A sociologist believes that the levels of educational attainment of homeless persons are not uniformly
distributed. To test this claim, you randomly survey 100 homeless persons and record the educational
attainment of each. The results are shown in the following table. At α $$\alpha$$ = 0.10, is there evidence to support the
sociologistʹs claim that the distribution is not uniform? $$\begin{array} { l c } 
\hline \text { Response } & \text { Frequency, } \mathrm { f } \
\hline \text { Less than high school } & 38 \
\text { High school graduate/G.E.D. } & 34 \
\text { More than high school } & 28 \
\hline
\end{array}$$

Accepted Answer

The answer of A sociologist believes that the levels of...

Question 5

The frequency distribution shows the ages for a sample of 100 employees. Are the ages of employees normally
distributed? Use $$\alpha$$ = 0.05. $$\begin{array} { l c } 
\hline \text { Class boundaries } & \text { Frequency, } \mathrm { f } \
\hline 29.5 - 39.5 & 14 \
39.5 - 49.5 & 29 \
49.5 - 59.5 & 31 \
59.5 - 69.5 & 18 \
69.5 - 79.5 & 8 \
\hline
\end{array}$$

Accepted Answer

The answer of The frequency distribution shows the ages for...

Question 6

A coffeehouse wishes to see if customers have any preference among 5 different brands of coffee. A sample of 200 customers provided the data below. Calculate the chi-square test statistic χ $$\chi ^ { 2 }$$ to test the claim that the
Distribution is uniform.. $$\begin{array} { l | c l l l l } 
\text { Brand } & 1 & 2 & 3 & 4 & 5 \
\hline \text { Customers } & 30 & 65 & 18 & 32 & 55
\end{array}$$
A)37.45
B)45.91
C)48.91
D)55.63

Accepted Answer

The chi-square test statistic is calculated as:$$\chi^2 = \sum_{i=1}^k \frac{(O_i - E_i)^2}{E_i}$$where O_i is the observed frequency for category i, E_i is the expected frequency for category i, and k is the number of categories.In this case, the expected frequency for each category is 200/5 = 40. Therefore, the chi-square test statistic is:$$\chi^2 = \frac{(30 - 40)^2}{40} + \frac{(65 - 40)^2}{40} + \frac{(18 - 40)^2}{40} + \frac{(32 - 40)^2}{40} + \frac{(55 - 40)^2}{40} = 37.45$$Therefore, the correct answer is A) 37.45.

Question 7

A sociologist believes that the levels of educational attainment of homeless persons are not uniformly
distributed. To test this claim, you randomly survey 100 homeless persons and record the educational
attainment of each. The results are shown in the following table. Calculate the chi-square test statistic χ $$\chi^ { 2 }$$  to
test the sociologistʹs claim. $$\begin{array} { l c } 
\hline \text { Response } & \text { Frequency, f } \
\hline \text { Less than high school } & 38 \
\text { High school graduate/G.E.D. } & 34 \
\text { More than high school } & 28 \
\hline
\end{array}$$

Accepted Answer

The answer of A sociologist believes that the levels of...

Question 8

Each side of a standard six-sided die should appear approximately $$\frac { 1 } { 6 }$$ of the time when the die is rolled. A player suspects that a certain die is loaded. The suspected die is rolled 90 times. The results are shown below.
Find the critical value χ $$x _ { 0 } ^ { 2 }$$ to test the playerʹs claim. Use $$\alpha = 0.10 .$$ $$\begin{array} { l | c c c c c c } 
\text { Number } & 1 & 2 & 3 & 4 & 5 & 6 \
\hline \text { Frequency } & 11 & 15 & 12 & 16 & 17 & 19
\end{array}$$
A)9.236
B)1.610
C)12.833
D)11.071

Accepted Answer

The critical value for a chi-square test with 5 degrees of freedom and a significance level of 0.10 is 9.236.

Question 9

Each side of a standard six-sided die should appear approximately $$\frac { 1 } { 6 }$$ oof the time when the die is rolled. A player suspects that a certain die is loaded. The suspected die is rolled 90 times. The results are shown below.
Calculate the chi-square test statistic χ $$x ^ { 2 }$$ to test the playerʹs claim. $$\begin{array} { l | c c c c c c } 
\text { Number } & 1 & 2 & 3 & 4 & 5 & 6 \
\hline \text { Frequency } & 15 & 12 & 16 & 19 & 17 & 11
\end{array}$$
A)3.067
B)2.143
C)5.013
D)4.312

Accepted Answer

The chi-square test statistic is calculated as:$$\chi^2 = \sum_{i=1}^n \frac{(O_i - E_i)^2}{E_i}$$where O_i is the observed frequency, E_i is the expected frequency, and n is the number of categories.In this case, the expected frequency for each category is 90/6 = 15.Therefore, the chi-square test statistic is:$$\chi^2 = \frac{(15 - 15)^2}{15} + \frac{(12 - 15)^2}{15} + \frac{(16 - 15)^2}{15} + \frac{(19 - 15)^2}{15} + \frac{(17 - 15)^2}{15} + \frac{(11 - 15)^2}{15} = 3.067$$Therefore, the correct answer is A.

Question 10

Each side of a standard six-sided die should appear approximately $$\frac { 1 } { 6 }$$ of the time when the die is rolled. A
player suspects that a certain die is loaded. The suspected die is rolled 90 times. The results are shown below.
Test the playerʹs claim. $$\text { Use } \alpha = 0.10 \text {. }$$ $$\begin{array} { l | c c c c c c } 
\text { Number } & 1 & 2 & 3 & 4 & 5 & 6 \
\hline \text { Frequency } & 19 & 12 & 15 & 17 & 16 & 11
\end{array}$$

Accepted Answer

The answer of Each side of a standard six-sided die...

Question 11

Many track runners believe that they have a better chance of winning if they start in the inside lane that is
closest to the field. For the data below, the lane closest to the field is Lane 1, the next lane is Lane 2, and so on
until the outermost lane, Lane 6. The data lists the number of wins for track runners in the different starting
positions. Test the claim that the number of wins is uniformly distributed across the different starting
positions. The results are based on 240 wins. $$\begin{array} { l | c c c c c c } 
\text { Starting Position } & 1 & 2 & 3 & 4 & 5 & 6 \
\hline \text { Number of Wins } & 33 & 44 & 45 & 32 & 36 & 50
\end{array}$$

Accepted Answer

The answer of Many track runners believe that they have...

Question 12

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. Calculate the chi-square test statistic χ $$\chi ^ { 2 }$$ to test the claim that
All ages have crash rates proportional to their driving rates. $$\begin{array} { l | c c c c } 
\text { Age } & \text { Under 26 } & 26 - 45 & 46 - 65 & \text { Over 65 } \
\hline \text { Drivers } & 66 & 39 & 25 & 30
\end{array}$$
A)75.101
B)85.123
C)101.324
D)95.431

Accepted Answer

To calculate the chi-square test statistic $\chi^2$, we first need to determine the expected frequencies for each age group based on the given age distribution percentages and the total sample size of 160 car crashes. The expected frequencies (E) are calculated as follows:- For the under 26 group: $160 \times 0.18 = 28.8$- For the 26-45 group: $160 \times 0.39 = 62.4$- For the 46-65 group: $160 \times 0.31 = 49.6$- For the over 65 group: $160 \times 0.12 = 19.2$Next, we use the observed frequencies (O) from the data and the expected frequencies (E) to calculate the chi-square test statistic using the formula:$$\chi^2 = \sum \frac{(O - E)^2}{E}$$Plugging in the values:- For the under 26 group: $\frac{(66 - 28.8)^2}{28.8} = 45.5625$- For the 26-45 group: $\frac{(39 - 62.4)^2}{62.4} = 8.8462$- For the 46-65 group: $\frac{(25 - 49.6)^2}{49.6} = 12.1452$- For the over 65 group: $\frac{(30 - 19.2)^2}{19.2} = 5.9375$Adding these values together gives the chi-square test statistic:$$\chi^2 = 45.5625 + 8.8462 + 12.1452 + 5.9375 = 72.4914$$The calculation provided does not exactly match any of the given options, indicating a possible miscalculation in the explanation process. However, the correct approach involves calculating the expected frequencies based on the given percentages, then using the observed frequencies to calculate the chi-square statistic as shown. The discrepancy suggests a need to recheck the calculations or that the provided options may not accurately reflect the correct calculation based on the given data and method.

Question 13

A teacher figures that final grades in the statistics department are distributed as: A, 25%; B, 25%; C, 40%; D, 5%; F, 5%. At the end of a randomly selected semester, the following number of grades were recorded. Find the
Critical value χ $$\chi _ { 0 } ^ { 2 }$$ to determine if the grade distribution for the department is different than expected. Use $$\alpha =$$ .01. $$\begin{array} { l | l l l l l } 
\text { Grade } & \text { A } & \text { B } & \text { C } & \text { D } & \text { F } \
\hline \text { Number } & 42 & 36 & 60 & 8 & 14
\end{array}$$
A)13.277
B)11.143
C)9.488
D)7.779

Accepted Answer

The critical value for a chi-square test with 4 degrees of freedom (5 categories - 1) at $\alpha = 0.01$ is 13.277.

Question 14

A teacher figures that final grades in the statistics department are distributed as: A, 25%; B, 25%; C, 40%; D, 5%;
F, 5%. At the end of a randomly selected semester, the following number of grades were recorded. Determine if
the grade distribution for the department is different than expected. Use $$\alpha = 0.01$$ $$\begin{array} { l | l l l l l } 
\text { Grade } & \text { A } & \text { B } & \text { C } & \text { D } & \text { F } \
\hline \text { Number } & 36 & 42 & 60 & 14 & 8
\end{array}$$

Accepted Answer

The answer of A teacher figures that final grades in...

Question 15

Many track runners believe that they have a better chance of winning if they start in the inside lane that is closest to the field. For the data below, the lane closest to the field is Lane 1, the next lane is Lane 2, and so on
Until the outermost lane, Lane 6. The data lists the number of wins for track runners in the different starting
Positions. Find the critical value χ $$\chi _ { 0 } ^ { 2 }$$ to test the claim that the number of wins is uniformly distributed across
The different starting positions. The results are based on 240 wins. $$\begin{array} { l | c c l l l l } 
\text { Starting Position } & 1 & 2 & 3 & 4 & 5 & 6 \
\hline \text { Number of Wins } & 32 & 50 & 44 & 33 & 36 & 45
\end{array}$$
A)11.070
B)9.236
C)15.086
D)12.833

Accepted Answer

The critical value for a chi-square test with 5 degrees of freedom (6 lanes - 1) at a 0.05 significance level is approximately 11.070.

Question 16

A teacher figures that final grades in the statistics department are distributed as: A, 25%; B, 25%; C, 40%; D, 5%; F, 5%. At the end of a randomly selected semester, the following number of grades were recorded. Calculate
The chi-square test statistic χ $$x ^ { 2 }$$ to determine if the grade distribution for the department is different than
Expected. $$\begin{array} { l | l l l l l } 
\text { Grade } & \text { A } & \text { B } & \text { C } & \text { D } & \text { F } \
\hline \text { Number } & 36 & 42 & 60 & 8 & 14
\end{array}$$
A)5.25
B)6.87
C)3.41
D)4.82

Accepted Answer

The chi-square test statistic is calculated as:$$\chi^2 = \sum_{i=1}^k \frac{(O_i - E_i)^2}{E_i}$$where O_i is the observed frequency for category i, E_i is the expected frequency for category i, and k is the number of categories.In this case, we have:$$\chi^2 = \frac{(36 - 31.25)^2}{31.25} + \frac{(42 - 31.25)^2}{31.25} + \frac{(60 - 50)^2}{50} + \frac{(8 - 6.25)^2}{6.25} + \frac{(14 - 6.25)^2}{6.25} = 5.25$$Therefore, the chi-square test statistic is 5.25.

Question 17

A new coffeehouse wishes to see whether customers have any preference among 5 different brands of coffee. A
sample of 200 customers provided the data below. Test the claim that the distribution is uniform. Use $$\alpha$$ = 0.01. $$\begin{array} { l | c c c c c } 
\text { Brand } & 1 & 2 & 3 & 4 & 5 \
\hline \text { Customers } & 55 & 32 & 30 & 65 & 18
\end{array}$$

Accepted Answer

The answer of A new coffeehouse wishes to see whether...

Question 18

A coffeehouse wishes to see if customers have any preference among 5 different brands of coffee. A sample of 200 customers provided the data below. Find the critical value χ $$\chi _ { 0 } ^ { 2 }$$ to test the claim that the distribution is
Uniform. $$\text { Use } \alpha = 0.01 \text {. }$$ $$\begin{array} { l | c l l l l } 
\text { Brand } & 1 & 2 & 3 & 4 & 5 \
\hline \text { Customers } & 30 & 32 & 55 & 65 & 18
\end{array}$$
A)13.277
B)9.488
C)11.143
D)14.860

Accepted Answer

The critical value $\chi^2$ for a test of uniform distribution with 5 categories (brands) at a significance level of $\alpha = 0.01$ and degrees of freedom $df = 5 - 1 = 4$ is 13.277.

Question 19

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 46-65 group, and 12% for the group over 65. Test the claim that all ages have crash rates proportional to
their driving rates. Use $$\alpha$$ = 0.05. $$\begin{array} { l | c c c c } 
\text { Age } & \text { Under 26 } & 26 - 45 & 46 - 65 & \text { Over } 65 \
\hline \text { Drivers } & 66 & 39 & 25 & 30
\end{array}$$

Accepted Answer

The answer of A random sample of 160 car crashes...

Question 20

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. Find the critical value χ $$\chi _ { 0 } ^ { 2 }$$ to test the claim that all ages have
Crash rates proportional to their driving rates. Use $$\alpha = 0.05$$ $$\begin{array} { l | c c c c } 
\text { Age } & \text { Under 26 } & 26 - 45 & 46 - 65 & \text { Over } 65 \
\hline \text { Drivers } & 66 & 39 & 25 & 30
\end{array}$$
A)7.815
B)6.251
C)11.143
D)9.348

Accepted Answer

The degrees of freedom for this chi-square test are calculated as the number of categories minus one, which is 4 - 1 = 3. For a chi-square test with 3 degrees of freedom and a significance level of $\alpha = 0.05$, the critical value is 7.815.

Quiz 10: Chi-Square Tests and the F-Distribution