Question 1

Suspecting that a die may be unfair,you want to investigate.To check,you roll it 54 times,recording the number of times each face appears. $$\begin{array} { l | l } 
\text { Face } & \text { Count } \
\hline 1 & 9 \
2 & 15 \
3 & 6 \
4 & 12 \
5 & 6 \
6 & 6
\end{array}$$ How many degrees of freedom are there?
A)12
B)6
C)3
D)5
E)16

Accepted Answer

For a chi-square test on a single die, the degrees of freedom are calculated as the number of categories (in this case, the faces of the die) minus 1. Since there are 6 faces, the degrees of freedom are 6 - 1 = 5.

Question 2

Suspecting that a die may be unfair,you want to investigate.To check,you roll it 54 times,recording the number of times each face appears. $$\begin{array} { l | l } 
\text { Face } & \text { Count } \
\hline 1 & 9 \
2 & 15 \
3 & 6 \
4 & 12 \
5 & 6 \
6 & 6
\end{array}$$ Find $x ^ { 2 }$ .
A)1.000
B)2.000
C)4.000
D)8.000
E)10.000

Accepted Answer

The chi-square ($x^2$) statistic is used to test the hypothesis that an observed frequency distribution fits a expected distribution. In this case, we expect each face of the die to appear an equal number of times if the die is fair. With 54 rolls, the expected frequency for each face is $54 \div 6 = 9$.The formula for $x^2$ is $\sum \frac{(O_i - E_i)^2}{E_i}$, where $O_i$ is the observed frequency and $E_i$ is the expected frequency for each category.- For face 1: $(9 - 9)^2 / 9 = 0$- For face 2: $(15 - 9)^2 / 9 = 36 / 9 = 4$- For face 3: $(6 - 9)^2 / 9 = 9 / 9 = 1$- For face 4: $(12 - 9)^2 / 9 = 9 / 9 = 1$- For face 5: $(6 - 9)^2 / 9 = 9 / 9 = 1$- For face 6: $(6 - 9)^2 / 9 = 9 / 9 = 1$Adding these up gives $x^2 = 0 + 4 + 1 + 1 + 1 + 1 = 8$.

Question 3

In studying the responses to a multiple-choice test question,the following sample data were obtained.Are the responses uniformly distributed? Use a 10% level of significance. $$\begin{array} { r | r r r r r } 
\text { Response } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } \
\hline \text { Frequency } & 12 & 15 & 16 & 18 & 19
\end{array}$$

Accepted Answer

H0: The responses are uniformly distributed. HA: The responses are not uniformly distributed. Test statistic: χ2 = 1.875 P-value = 0.7587 Fail to reject the null hypothesis.There is not sufficient evidence to warrant rejection of the claim that the responses occur with the same frequency.

Question 4

A candy company claims that its bags of mixed suckers are 20% strawberry,30% cherry,15% apple,10% lemon,and 25% grape.A bag was purchased,and the number of each type of flavour was recorded in the chart below. $$\begin{array} { l | c } 
{ \text { Flavor } } & \text { Count } \
\hline \text { Strawberry } & 31 \
\text { Cherry } & 15 \
\text { Apple } & 17 \
\text { Lemon } & 15 \
\text { Grape } & 22
\end{array}$$
A)H0: The distribution of flavours is not the same as the distribution claimed by the company. HA: The distribution of flavours is the same as the distribution claimed by the company.
B)H0: The distribution of flavours is the same as the distribution claimed by the company. HA: The distribution of flavours is not the same as the distribution claimed by the company.
C)H0: The distribution of flavours is not uniform. HA: The distribution of flavours is uniform.
D)H0: The distribution of flavours is uniform. HA: The distribution of flavours is not uniform.
E)H0: The flavours and counts are independent. HA: The flavours and counts are not independent.

Accepted Answer

The null hypothesis (H0) should always assume no effect or difference, so it should state that the distribution is the same as claimed by the company. Therefore, option B is the correct option. The alternative hypothesis (HA) should state that the distribution is different from what the company claims.

Question 5

Test the claim that the responses occur with percentages of 15%,20%,25%,25%,and 15% respectively.Use a 10% level of significance. $$\begin{array} { r | r r r r r } 
\text { Response } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } \
\hline \text { Frequency } & 12 & 15 & 16 & 18 & 19
\end{array}$$

Accepted Answer

The answer of Test the claim that the responses occur...

Question 6

A study conducted for a woman's magazine indicates that the number of children desired by newly-married Canadian couples is distributed as follows.
Number of children desired Percent
0 9%
1 17%
2 55%
3 11%
Over 3 8%
A random sample of 600 couples in Ontario resulted in the following data.
Number of children desired Frequency
0 55
1 118
2 340
3 60
Over 3 27
Ontario couples is different from that of the Canadian population as a whole.Use a 5% level of significance.

Accepted Answer

The answer of A study conducted for a woman's magazine...

Question 7

You roll a die 48 times with the following results.Is the die fair? $$\begin{array} { c | r | r | r | r | r | r } 
\text { Number } & 1 & 2 & 3 & 4 & 5 & 6 \
\hline \text { Frequency } & 14 & 13 & 2 & 12 & 3 & 4
\end{array}$$
A)H0: The die is not fair. HA: The die is fair (all numbers occur with equal frequency).
Conclusion: Reject the null hypothesis.There is sufficient evidence to claim that the numbers on the die do not occur with equal frequency.
B)H0: The die is not fair. HA: The die is fair (all numbers occur with equal frequency).
Conclusion: Do not reject the null hypothesis.There is sufficient evidence to claim that the numbers on the die do not occur with equal frequency.
C)H0: The die is fair (all numbers occur with equal frequency). HA: The die is not fair.
Conclusion: Do not reject the null hypothesis.There is insufficient evidence to claim that the numbers on the die do not occur with equal frequency.
D)H0: The die is fair (all numbers occur with equal frequency). HA: The die is not fair.
Conclusion: Reject the null hypothesis.There is sufficient evidence to claim that the numbers on the die do not occur with equal frequency.
E)H0: The die is not fair. HA: The die is fair (all numbers occur with equal frequency).
Conclusion: Do not reject the null hypothesis.There is insufficient evidence to claim that the numbers on the die do not occur with equal frequency.

Accepted Answer

The hypothesis test should be set up as H0: The die is fair (all numbers occur with equal frequency) and HA: The die is not fair. Then, a chi-squared goodness-of-fit test can be conducted using the observed frequencies and expected frequencies (which would be 8 for each number) to determine if there is enough evidence to reject the null hypothesis. If the test statistic exceeds the critical value, the null hypothesis is rejected, indicating that the die is not fair. The correct answer is choice D, which states that the null hypothesis is rejected and there is sufficient evidence to claim that the numbers on the die do not occur with equal frequency.

Question 8

A die is rolled 180 times and the following data are obtained.Is the die fair? Number Frequency
1 31
2 34
3 26
4 16
5 32
6 41
A)H0: The die is not fair. HA: The die is fair (all numbers occur with equal frequency).
Conclusion: Do not reject the null hypothesis.There is insufficient evidence to claim that the numbers on the die do not occur with equal frequency.
B)H0: The die is not fair. HA: The die is fair (all numbers occur with equal frequency).
Conclusion: Do not reject the null hypothesis.There is sufficient evidence to claim that the numbers on the die do not occur with equal frequency.
C)H0: The die is fair (all numbers occur with equal frequency). HA: The die is not fair.
Conclusion: Do not reject the null hypothesis.The data do not provide sufficient evidence to conclude that the die is loaded.
D)H0: The die is fair (all numbers occur with equal frequency). HA: The die is not fair.
Conclusion: Reject the null hypothesis.There is sufficient evidence to claim that the numbers on the die do not occur with equal frequency.
E)H0: The die is not fair. HA: The die is fair (all numbers occur with equal frequency).
Conclusion: Reject the null hypothesis.There is sufficient evidence to claim that the numbers on the die do not occur with equal frequency.

Accepted Answer

The correct hypothesis setup for testing fairness is to assume fairness as the null hypothesis (H0: the die is fair) and unfairness as the alternative hypothesis (HA: the die is not fair). The given frequencies show significant variation from the expected frequency (180/6 = 30 for each number if the die were fair), suggesting the die may not be fair, leading to the rejection of the null hypothesis.

Question 9

Test the claim that workplace accidents are distributed on workdays as follows: Monday 25%,Tuesday: 15%,Wednesday: 15%,Thursday: 15%,and Friday: 30%.
In a study of 100 workplace accidents,24 occurred on a Monday,15 occurred on a Tuesday,17 occurred on a Wednesday,18 occurred on a Thursday,and 26 occurred on a Friday.Use a 10% level of significance.

Accepted Answer

The answer of Test the claim that workplace accidents are...

Question 10

Suspecting that a die may be unfair,you want to investigate.To check,you roll it 18 times,recording the number of times each face appears. $$\begin{array} { l | l } 
\text { Face } & \text { Count } \
\hline 1 & 3 \
2 & 5 \
3 & 2 \
4 & 4 \
5 & 2 \
6 & 2
\end{array}$$ If the die is fair,how many times would you expect each face to show?
A)2
B)6
C)3
D)4
E)1

Accepted Answer

If the die is fair, each face should appear an equal number of times. Since the die is rolled 18 times and there are 6 faces, each face should appear 18/6 = 3 times.

Question 11

Explain why a chi-square goodness-of-fit test is always right-tailed.

Accepted Answer

The answer of Explain why a chi-square goodness-of-fit test is...

Question 12

A die,suspected of being unfair,was rolled 50 times.The number of times each face appeared was recorded in the following table. $$\begin{array} { c | c } 
\text { Face } & \text { Count } \
\hline 1 & 8 \
2 & 10 \
3 & 9 \
4 & 11 \
5 & 5 \
6 & 7
\end{array}$$
A)H0: The counts are normally distributed over the faces. HA: The counts are not normally distributed over the faces.
B)H0: The counts and faces are independent. HA: The counts and faces are not independent.
C)H0: The counts and faces are not independent. HA: The counts and faces are independent.
D)H0: The counts are not uniformly distributed over the faces. HA: The counts are uniformly distributed over the faces.
E)H0: The counts are uniformly distributed over the faces. HA: The counts are not uniformly distributed over the faces.

Accepted Answer

The null hypothesis (H0) should be that the counts are uniformly distributed over the faces, and the alternative hypothesis (HA) should be that the counts are not uniformly distributed, which aligns with option E.

Question 13

A company manager wishes to test a union leader's claim that absences occur on the different week days with the same frequencies.Are absences uniformly distributed over the week? Use a 1% level of significance. $$\begin{array} { c | r r r r r } 
\text { Day } & \text { Mon } & \text { Tue } & \text { Wed } & \text { Thur } & \text { Fri } \
\hline \text { Absences } & 37 & 15 & 12 & 23 & 43
\end{array}$$

Accepted Answer

The answer of A company manager wishes to test a...

Question 14

When performing a chi-square goodness-of-fit test,why do large values of the test statistic suggest that the variable does not have the distribution stated in the null hypothesis? Refer to the formula for the test statistic in your explanation.

Accepted Answer

The answer of When performing a chi-square goodness-of-fit test,why do...

Question 15

Suspecting that a die may be unfair,you want to investigate.To check,you roll it 90 times,recording the number of times each face appears. $$\begin{array} { l | l } 
\text { Face } & \text { Count } \
\hline 1 & 15 \
2 & 25 \
3 & 10 \
4 & 20 \
5 & 10 \
6 & 10
\end{array}$$ State your hypotheses.
A)H0: The die is not fair. HA: The die is fair (all faces have p = 1/6).
B)H0: The die is not fair. HA: The die is fair (all faces have p = 1/2).
C)H0: The die is fair (all faces have p = 1/6). HA: The die is not fair.
D)H0: The die is fair (all faces have p = 1/3). HA: The die is not fair.
E)H0: The die is not fair. HA: The die is fair (all faces have p = 1/3).

Accepted Answer

The null hypothesis (H0) should always be the assumption that is tested against the alternative hypothesis (HA). In this case, the null hypothesis should be that the die is fair (all faces have equal probability of showing up). The alternative hypothesis can be either that the die is not fair or that it has specific biased probabilities for certain faces. Option C is the correct choice because it assumes that the die is fair (H0: all faces have p=1/6) and the alternative hypothesis is that it is not. The other options are not appropriate because they either assume the die is biased to some specific probability or the null and alternative hypotheses are flipped.

Question 16

Suspecting that a die may be unfair,you want to investigate.To check,you roll it 36 times,recording the number of times each face appears. $$\begin{array} { l | l } 
\text { Face } & \text { Count } \
\hline 1 & 6 \
2 & 10 \
3 & 4 \
4 & 8 \
5 & 4 \
6 & 4
\end{array}$$ To see if these results are unusual,will you test goodness-of-fit,homogeneity,or independence?
A)Independence
B)Goodness-of-fit
C)Homogeneity

Accepted Answer

The best choice for this scenario is to test the goodness-of-fit. Goodness-of-fit tests are used to determine if a given sample fits a particular distribution. In this case, we want to see if the results of rolling the die fit the expected distribution of a fair die. The chi-squared goodness-of-fit test can be used to compare the observed frequencies (number of times each face appears) with the expected frequencies (assuming a fair die) and determine if they are significantly different. Therefore, the correct answer is B.

Question 17

In studying the occurrence of genetic characteristics,the following sample data were obtained.Do the characteristics occur with the same frequency? Use a 10% level of significance. $$\begin{array} { c | r r r r r r } 
\text { Characteristic } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } & \text { F } \
\hline \text { Frequency } & 28 & 30 & 45 & 48 & 38 & 39
\end{array}$$ χ2 = 8.263; P-value = 0.1423
A)Reject the null hypothesis.There is sufficient evidence to warrant rejection of the claim that the characteristics occur with the same frequency.
B)Fail to reject the null hypothesis.There is not sufficient evidence to warrant rejection of the claim that the characteristics occur with the same frequency.
C)Fail to reject the null hypothesis.There is not sufficient evidence to warrant rejection of the claim that the characteristics are independent.
D)Fail to reject the null hypothesis.There is sufficient evidence to warrant rejection of the claim that the characteristics occur with the same frequency.
E)Reject the null hypothesis.There is not sufficient evidence to warrant rejection of the claim that the characteristics occur with the same frequency.

Accepted Answer

The P-value is greater than the significance level of 0.1, which means we fail to reject the null hypothesis. Hence, there is not sufficient evidence to warrant rejection of the claim that the characteristics occur with the same frequency.

Question 18

Among the eastern provinces in 2010,states,Nova Scotia has 40% of the total population,New Brunswick has 33%,Newfoundland has 21%,and Prince Edward Island has 6%.A market researcher selects a sample of 1000 subjects in 2011,with 450 in Nova Scotia,340 in New Brunswick,150 in Newfoundland,and 60 in Prince Edward Island.Does the distribution of the sample agree with the population distribution? Use a 1% level of significance.

Accepted Answer

The answer of Among the eastern provinces in 2010,states,Nova Scotia...

Question 19

A die is rolled 180 times and the following data are obtained. Number Frequency
1 31
2 34
3 26
4 16
5 32
6 41
A chi-square test was conducted to determine whether or not the die is loaded (i.e.,that the six numbers are not equally likely).Use a 5% level of significance. $x ^ { 2 }$ = 11.8; P-value = 0.03763
A)Reject the null hypothesis.The data provide sufficient evidence to conclude that the die is loaded.
B)Reject the null hypothesis.The data provide sufficient evidence to conclude that the die is fair.
C)Do not reject the null hypothesis.The data provide sufficient evidence to conclude that the die is loaded.
D)Do not reject the null hypothesis.The data do not provide sufficient evidence to conclude that the die is loaded.
E)Reject the null hypothesis.The data do not provide sufficient evidence to conclude that the die is loaded.

Accepted Answer

The null hypothesis is that the die is fair (i.e., the six numbers are equally likely). The alternative hypothesis is that the die is loaded (i.e., the six numbers are not equally likely). The chi-square test statistic is 11.8, and the corresponding p-value is 0.03763. Since the p-value is less than the significance level of 0.05, we reject the null hypothesis. Therefore, the data provide sufficient evidence to conclude that the die is loaded.

Question 20

In studying the occurrence of genetic characteristics,the following sample data were obtained.Are they uniformly distributed? Use a 10% level of significance. $$\begin{array} { c | r r r r r r } 
\text { Characteristic } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } & \text { F } \
\hline \text { Frequency } & 28 & 30 & 45 & 48 & 38 & 39
\end{array}$$

Accepted Answer

The answer of In studying the occurrence of genetic characteristics,the...

Quiz 23: Comparing Counts

Suspecting that a die may be unfair,you want to investigate.To check,you roll it 54 times,recording the number of times each face appears. $\begin{array} { l | l } \text { Face } & \text { Count } \\\hline 1 & 9 \\2 & 15 \\3 & 6 \\4 & 12 \\5 & 6 \\6 & 6\end{array}$ Find $x ^ { 2 }$ .

You roll a die 48 times with the following results.Is the die fair? $\begin{array} { c | r | r | r | r | r | r } \text { Number } & 1 & 2 & 3 & 4 & 5 & 6 \\\hline \text { Frequency } & 14 & 13 & 2 & 12 & 3 & 4\end{array}$

A die is rolled 180 times and the following data are obtained.Is the die fair? Number Frequency 1 31 2 34 3 26 4 16 5 32 6 41

Explain why a chi-square goodness-of-fit test is always right-tailed.

A die,suspected of being unfair,was rolled 50 times.The number of times each face appeared was recorded in the following table. $\begin{array} { c | c } \text { Face } & \text { Count } \\\hline 1 & 8 \\2 & 10 \\3 & 9 \\4 & 11 \\5 & 5 \\6 & 7\end{array}$

When performing a chi-square goodness-of-fit test,why do large values of the test statistic suggest that the variable does not have the distribution stated in the null hypothesis? Refer to the formula for the test statistic in your explanation.

Suspecting that a die may be unfair,you want to investigate.To check,you roll it 90 times,recording the number of times each face appears. $\begin{array} { l | l } \text { Face } & \text { Count } \\\hline 1 & 15 \\2 & 25 \\3 & 10 \\4 & 20 \\5 & 10 \\6 & 10\end{array}$ State your hypotheses.