Question 1

The following table shows the Myers-Briggs personality preferences for a random sample of 401 people in the listed professions. Use the chi-square test to determine if the listed occupations and personality preferences are independent at . What is the level of significance?
$$\begin{array} { l | l | l | l } 
\text { Occupation } & \text { Extroverted } & \text { Introverted } & \text { Row Total } \
\hline \text { Clergy (all denominations) } & 60 & 48 & 108 \
\hline \text { M.D } & 65 & 94 & 159 \
\hline \text { Lawyer } & 54 & 80 & 134 \
\hline \text { Column Total } & 179 & 222 & 401 \
\hline
\end{array}$$ $$\alpha = 0.05$$
A)0.99
B)0.05
C)0.10
D)0.90
E)0.95

Accepted Answer

The level of significance, denoted as $\alpha$, is given in the question as 0.05.

Question 2

A transect is an archaeological study area that is mile wide and 1 mile long. A site in a transect is the location of a significant archaeological find. Let x represent the number of sites per transect. In a section of Chaco Canyon, a large number of transects showed that x has a population variance In a different section of Chaco Canyon, a random sample of 24 transects gave a sample variance for the number of sites per transect. Use an to test the claim that the variance in the new section is greater than 35.2. $\frac { 1 } { 5 }$ $\sigma ^ { 2 } = 35.2$ $s ^ { 2 } = 58.1$ $\alpha = 0.1$ What is the level of significance?
A)95%
B)1%
C)90%
D)99%
E)10%

Accepted Answer

The level of significance is given as $\alpha = 0.1$, which corresponds to 10%.

Question 3

Suppose the age distribution of the Canadian population and the age distribution of a random sample of 531 residents in the Indian community of Red Lake are shown below. $$\begin{array} { l l l } 
& & \text { Observed Number } \
\text { Age (years) } & \text { Percent of Canadian Population } & \text { in Red Lake Village } \
\hline \text { Under } 5 & 8.9 \% & 49 \
5 \text { to } 14 & 11.9 \% & 41 \
15 \text { to } 64 & 67.8 \% & 389 \
65 \text { and older } & 11.4 \% & 52 \
\hline
\end{array}$$ Use to test the claim that the age distribution of the general Canadian population fits the age distribution of the residents of Red Lake Village. What are the degrees of freedom?
$$\alpha = 0.05$$
A)7
B)3
C)5
D)4
E)6

Accepted Answer

The degrees of freedom for a chi-square test are calculated as the number of categories minus one. Here, there are 4 age categories, so the degrees of freedom are $4 - 1 = 3$.

Question 4

Suppose the age distribution of the Canadian population and the age distribution of a random sample of 447 residents in the Indian community of Red Lake are shown below. Use to test the claim that the age distribution of the general Canadian population fits the age distribution of the residents of Red Lake Village. State the null and alternate hypotheses.
$$\begin{array} { l l l } 
\hline \text { Age (years) } & \text { Percent of Canadian Population } & \begin{array} { l } 
\text { Observed Number } \
\text { in Red Lake Village }
\end{array} \
\hline \text { Under } 5 & 7.9 \% & 45 \
5 \text { to } 14 & 13.3 \% & 55 \
15 \text { to } 64 & 65.3 \% & 302 \
65 \text { and older } & 13.5 \% & 45 \
\hline
\end{array}$$ $\alpha = 0.1$
A)The distributions are equal to zero ; The distributions are not the same. $H _ { 0 } :$ $H _ { 1 } :$
B)The distributions are the same; The distributions are not the same. $H _ { 0 } :$ $H _ { 1 } :$
C)The distributions are equal to zero; The distributions are the same. $H _ { 0 } :$ $H _ { 1 }$
D)The distributions are different; The distributions are equal to zero. $H _ { 0 } :$ $H _ { 1 }$:
E)The distributions are not equal to zero; The distributions are not the same. $H _ { 0 } :$ $H _ { 1 }$:

Accepted Answer

The null hypothesis ($H_0$) typically states that there is no difference or no effect, so in this context, it would state that the age distribution of the general Canadian population is the same as the age distribution of the residents of Red Lake Village. The alternative hypothesis ($H_1$) is the opposite of the null and would claim that the distributions are not the same.

Question 5

The following table shows the Myers-Briggs personality preferences for a random sample of 402 people in the listed professions. $$\begin{array} { l | l | l | l } 
\text { Occupation } & \text { Extroverted } & \text { Introverted } & \text { Row Total } \
\hline \text { Clergy (all denominations) } & 60 & 46 & 106 \
\hline \text { M.D. } & 70 & 92 & 162 \
\hline \text { Lawyer } & 53 & 81 & 134 \
\hline \text { Column Total } & 183 & 219 & 402 \
\hline
\end{array}$$ Use the chi-square test to determine if the listed occupations and personality preferences are independent at . State the null and alternate hypotheses.
$\alpha = 0.01$
A)Myers-Briggs preference and profession are not independent; Myers-Briggs preference is independent whereas profession is not independent $H _ { 0 } :$ $H _ { 1 }$
B)Myers-Briggs preference and profession are independent; Myers-Briggs preference is not independent whereas profession is independent $H _ { 0 } :$ $H _ { 1 }$:
C)Myers-Briggs preference and profession are not independent; Myers-Briggs preference and profession are independent $H _ { 0 } :$ $H _ { 1 }$:
D)Myers-Briggs preference is independent whereas profession is not independent; Myers-Briggs preference is not independent whereas profession is independent $H _ { 0 } :$ $H _ { 1 }$:
E)Myers-Briggs preference and profession are independent; Myers-Briggs preference and profession are not independent $H _ { 0 } :$ $H _ { 1 }$:

Accepted Answer

The null hypothesis (H0) always states that there is no effect or no difference, so in this context, it would state that Myers-Briggs preference and profession are independent. The alternative hypothesis (H1) is the opposite, stating that Myers-Briggs preference and profession are not independent.

Question 6

The following table shows the Myers-Briggs personality preferences for a random sample of 404 people in the listed professions. $$\begin{array} { l | l | l | l } 
\text { Occupation } & \text { Extroverted } & \text { Introverted } & \text { Row Total } \
\hline \text { Clergy (all denominations) } & 60 & 45 & 105 \
\hline \text { M.D. } & 65 & 94 & 159 \
\hline \text { Lawyer } & 55 & 85 & 140 \
\hline \text { Column Total } & 180 & 224 & 404 \
\hline
\end{array}$$ Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.1 level of significance. Find (or estimate) the P-value of the sample test statistic.
A)0.005 < P-Value < 0.01
B)0.025 < P-Value < 0.05
C)P-Value > 0.5
D)0.01 < P-Value < 0.025
E)0.10 < P-Value < 0.25

Accepted Answer

The chi-square test statistic is calculated and compared to the chi-square distribution with the appropriate degrees of freedom. The P-value falls between 0.01 and 0.025, indicating a significant association between occupation and personality preference at the 0.1 level of significance.

Question 7

Suppose the age distribution of the Canadian population and the age distribution of a random sample of 492 residents in the Indian community of Red Lake are shown below. $$\begin{array} { l l l } 
& & \text { Observed Number } \
\text { Age (years) } & \text { Percent of Canadian Population } & \text { in Red Lake Village } \
\hline \text { Under } 5 & 8.5 \% & 33 \
5 \text { to } 14 & 10.3 \% & 38 \
15 \text { to } 64 & 70.5 \% & 373 \
65 \text { and older } & 10.7 \% & 48 \
\hline
\end{array}$$ Use to test the claim that the age distribution of the general Canadian population fits the age distribution of the residents of Red Lake Village. Given that 0.05 < P-Value < 0.10, will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories?
$$\alpha = 0.01$$
A)Since the P-Value is less than $\alpha$, we fail to reject the null hypothesis that the variables are independent. At 0.01 level of significance, we conclude that the variables are not independent.
B)Since the P-Value is less than $\alpha$, we reject the null hypothesis that the variables are independent. At 0.01 level of significance, we conclude that the variables are not independent.
C)Since the P-Value is less than $\alpha$, we fail to reject the null hypothesis that the variables are not independent. At 0.01 level of significance, we conclude that the variables are independent.
D)Since the P-Value is less than $\alpha$, we reject the null hypothesis that the variables are not independent. At 0.01 level of significance, we conclude that the variables are independent.
E)Since the P-Value is greater than $\alpha$, we fail to reject the null hypothesis that the variables are independent. At 0.01 level of significance, we conclude that the variables are not independent.

Accepted Answer

The answer of Suppose the age distribution of the Canadian...

Question 8

A transect is an archaeological study area that is mile wide and 1 mile long. A site in a transect is the location of a significant archaeological find. Let x represent the number of sites per transect. In a section of Chaco Canyon, a large number of transects showed that x has a population variance In a different section of Chaco Canyon, a random sample of 23 transects gave a sample variance for the number of sites per transect. Use an to test the claim that the variance in the new section is greater than 37.9. $\frac { 1 } { 5 }$ $\sigma ^ { 2 } = 37.9$ $s ^ { 2 } = 61.6$ $\alpha = 0.05$ State the null and alternate hypotheses. A); $H _ { 0 } :$ $\sigma ^ { 2 } > 37.9$ $H _ { 1 }$: $\sigma ^ { 2 } < 37.9$ B); $H _ { 0 } :$ $\sigma ^ { 2 } = 37.9$ $H _ { 1 }$: $\sigma ^ { 2 } \neq 37.9$ C); $H _ { 0 } :$ $\sigma ^ { 2 } \neq 37.9$ $H _ { 1 }$: $\sigma ^ { 2 } = 37.9$ D); $H _ { 0 } :$ $\sigma ^ { 2 } = 37.9$ $H _ { 1 }$: $\sigma ^ { 2 } > 37.9$ E); $H _ { 0 } :$ $\sigma ^ { 2 } = 37.9$ $H _ { 1 }$: $\sigma ^ { 2 } < 37.9$

Accepted Answer

The null hypothesis ($H_0$) always states that there is no effect or no difference, or it states the condition of equality. In this case, it would state that the variance is equal to 37.9. The alternative hypothesis ($H_1$) is what you are testing for, in this case, that the variance in the new section is greater than 37.9. Therefore, $H_0: \sigma^2 = 37.9$ and $H_1: \sigma^2 > 37.9$.

Question 9

The following table shows the Myers-Briggs personality preferences for a random sample of 399 people in the listed professions. $$\begin{array} { l | l | l | l } 
\text { Occupation } & \text { Extroverted } & \text { Introverted } & \text { Row Total } \
\hline \text { Clergy (all denominations) } & 64 & 44 & 108 \
\hline \text { M.D. } & 69 & 90 & 159 \
\hline \text { Lawyer } & 52 & 80 & 132 \
\hline \text { Column Total } & 185 & 214 & 399 \
\hline
\end{array}$$ Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.01 level of significance. Given P-Value < 0.005, will you reject or fail to reject the null hypothesis of independence?
A)Since the P-value is greater than $\alpha$, we fail to reject the null hypothesis that the Myers-Briggs personality preference and profession are not independent. At 0.01 level of significance, we conclude that the Myers-Briggs personality preference and profession are independent.
B)Since the P-value is greater than $\alpha$, we fail to reject the null hypothesis that the Myers-Briggs personality preference and profession are independent. At 0.01 level of significance, we conclude that the Myers-Briggs personality preference and profession are not independent.
C)Since the P-value is less than $\alpha$, we reject the null hypothesis that the Myers-Briggs personality preference and profession are independent. At 0.01 level of significance, we conclude that the Myers-Briggs personality preference and profession are not independent.
D)Since the P-value is greater than $\alpha$, we reject the null hypothesis that the Myers-Briggs personality preference and profession are not independent. At 0.01 level of significance, we conclude that the Myers-Briggs personality preference and profession are independent.
E)Since the P-value is less than $\alpha$, we fail to reject the null hypothesis that the Myers-Briggs personality preference and profession are not independent. At 0.01 level of significance, we conclude that the Myers-Briggs personality preference and profession are independent.

Accepted Answer

Since the P-value is less than 0.01, we reject the null hypothesis, indicating that there is a significant association between Myers-Briggs personality preference and profession, meaning they are not independent.

Question 10

A transect is an archaeological study area that is mile wide and 1 mile long. A site in a transect is the location of a significant archaeological find. Let x represent the number of sites per transect. In a section of Chaco Canyon, a large number of transects showed that x has a population variance In a different section of Chaco Canyon, a random sample of 24 transects gave a sample variance for the number of sites per transect. Use an to test the claim that the variance in the new section is greater than 42.1. $$\frac { 1 } { 5 }$$ $\sigma ^ { 2 } = 42.1$ $s ^ { 2 } = 57.7$ $\alpha = 0.1$ What are the degrees of freedom?
A)26
B)23
C)24
D)27
E)25

Accepted Answer

The degrees of freedom for a sample variance test are calculated as the sample size minus one. Given a sample size of 24 transects, the degrees of freedom (df) are 24 - 1 = 23.

Question 11

Suppose the age distribution of the Canadian population and the age distribution of a random sample of 454 residents in the Indian community of Red Lake are shown below. $$\begin{array} { l l l } 
& & \text { Observed Number } \
\text { Age (years) } & \text { Percent of Canadian Population } & \text { in Red Lake Village } \
\hline \text { Under } 5 & 8.9 \% & 34 \
5 \text { to } 14 & 14.1 \% & 58 \
15 \text { to } 64 & 62.4 \% & 302 \
65 \text { and older } & 14.6 \% & 60 \
\hline
\end{array}$$ Use to test the claim that the age distribution of the general Canadian population fits the age distribution of the residents of Red Lake Village. Will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories?
$$\alpha = 0.05$$
A)Since the P-Value is less than $\alpha$, we reject the null hypothesis that the variables are independent in favor of the alternate hypothesis that the variables are not independent.
B)Since the P-Value is greater than $\alpha$, we fail to reject the null hypothesis that the variables are not independent in favor of the alternate hypothesis that the variables are independent.
C)Since the P-Value is less than $\alpha$, we fail to reject the null hypothesis that the variables are independent in favor of the alternate hypothesis that the variables are not independent.
D)Since the P-Value is less than $\alpha$, we reject the null hypothesis that the variables are not independent in favor of the alternate hypothesis that the variables are independent.
E)Since the P-Value is greater than $\alpha$, we reject the null hypothesis that the variables are not independent in favor of the alternate hypothesis that the variables are independent.

Accepted Answer

The answer of Suppose the age distribution of the Canadian...

Question 12

The following table shows the Myers-Briggs personality preferences for a random sample of 400 people in the listed professions. $$\begin{array} { l | l | l | l } 
\text { Occupation } & \text { Extroverted } & \text { Introverted } & \text { Row Total } \
\hline \text { Clergy (all denominations) } & 63 & 42 & 105 \
\hline \text { M.D. } & 67 & 95 & 162 \
\hline \text { Lawyer } & 52 & 81 & 133 \
\hline \text { Column Total } & 182 & 218 & 400 \
\hline
\end{array}$$ Use the chi-square test to determine if the listed occupations and personality preferences are independent at . Find the value of the chi-square statistic for the sample.
$$\alpha = 0.01$$
A)12.22
B)0.01
C)12.01
D)0.22
E)9.20

Accepted Answer

To calculate the chi-square statistic, we first need to find the expected frequencies for each cell under the assumption that occupation and personality preference are independent. The expected frequency for a cell is calculated as $(\text{Row Total} \times \text{Column Total}) / \text{Total}$.For Clergy (Extroverted), the expected frequency is $(105 \times 182) / 400 = 47.715$.For Clergy (Introverted), it's $(105 \times 218) / 400 = 57.285$.For M.D. (Extroverted), it's $(162 \times 182) / 400 = 73.71$.For M.D. (Introverted), it's $(162 \times 218) / 400 = 88.29$.For Lawyer (Extroverted), it's $(133 \times 182) / 400 = 60.575$.For Lawyer (Introverted), it's $(133 \times 218) / 400 = 72.425$.Next, we calculate the chi-square statistic using the formula $\chi^2 = \sum \frac{(O - E)^2}{E}$, where $O$ is the observed frequency and $E$ is the expected frequency.For Clergy (Extroverted), $\chi^2 = \frac{(63 - 47.715)^2}{47.715}$.For Clergy (Introverted), $\chi^2 = \frac{(42 - 57.285)^2}{57.285}$.For M.D. (Extroverted), $\chi^2 = \frac{(67 - 73.71)^2}{73.71}$.For M.D. (Introverted), $\chi^2 = \frac{(95 - 88.29)^2}{88.29}$.For Lawyer (Extroverted), $\chi^2 = \frac{(52 - 60.575)^2}{60.575}$.For Lawyer (Introverted), $\chi^2 = \frac{(81 - 72.425)^2}{72.425}$.Adding all these values together gives us the total chi-square statistic, which is approximately 12.22.

Question 13

Suppose the age distribution of the Canadian population and the age distribution of a random sample of 406 residents in the Indian community of Red Lake are shown below. Use to test the claim that the age distribution of the general Canadian population fits the age distribution of the residents of Red Lake Village. Given a value of 10.369 for , find (or estimate) the P-value of the sample test statistic.
$$\begin{array} { l l l } 
& & \text { Observed Number } \
\text { Age (years) } & \text { Percent of Canadian Population } & \text { in Red Lake Village } \
\hline \text { Under } 5 & 5.9 \% & 13 \
5 \text { to } 14 & 10.6 \% & 31 \
15 \text { to } 64 & 72.4 \% & 318 \
65 \text { and older } & 11.1 \% & 44\
\hline
\end{array}$$ $$\alpha = 0.1$$ $$\chi ^ { 2 }$$
A)0.05 < P-Value < 0.10
B)0.10 < P-Value < 0.25
C)0.005 < P-Value < 0.01
D)0.01 < P-Value < 0.025
E)P-Value < 0.005

Accepted Answer

The answer of Suppose the age distribution of the Canadian...

Question 14

Suppose the age distribution of the Canadian population and the age distribution of a random sample of 452 residents in the Indian community of Red Lake are shown below. Use to test the claim that the age distribution of the general Canadian population fits the age distribution of the residents of Red Lake Village. Find (or estimate) the P-value of the sample test statistic.
$$\begin{array} { l l l } 
\hline \text { Age (years) } & \text { Percent of Canadian Population } & \begin{array} { l } 
\text { Observed Number } \
\text { in Red Lake Village }
\end{array} \
\hline \text { Under } 5 & 7.5 \% & 50 \
5 \text { to } 14 & 12.5 \% & 45 \
15 \text { to } 64 & 69.9 \% & 313 \
65 \text { and older } & 10.1 \% & 44 \
\hline
\end{array}$$ $$\alpha = 0.05$$
A)P-Value > 0.50
B)0.25 < P-Value < 0.50
C)0.005 < P-Value < 0.01
D)0.025 < P-Value < 0.05
E)0.01 < P-Value < 0.025

Accepted Answer

The answer of Suppose the age distribution of the Canadian...

Question 15

Suppose the age distribution of the Canadian population and the age distribution of a random sample of 505 residents in the Indian community of Red Lake are shown below. $$\begin{array} { l l l } 
\hline & & \text { Observed Number } \
\text { Age (years) } & \text { Percent of Canadian Population } & \text { in Red Lake Village } \
\hline \text { Under } 5 & 8.5 \% & 45 \
5 \text { to } 14 & 10.4 \% & 50 \
15 \text { to } 64 & 68.2 \% & 367 \
65 \text { and older } & 12.9 \% & 43 \
\hline
\end{array}$$ Use to test the claim that the age distribution of the general Canadian population fits the age distribution of the residents of Red Lake Village. Find the value of the chi-square statistic for the sample.
$$\alpha = 0.05$$
A)9.231
B)0.274
C)0.124
D)10.945
E)11.136

Accepted Answer

To find the chi-square statistic, we first calculate the expected number of residents in each age group based on the percentages of the Canadian population, then apply the formula for the chi-square statistic: $\chi^2 = \sum \frac{(O - E)^2}{E}$, where $O$ is the observed frequency and $E$ is the expected frequency.1. Calculate expected frequencies (E) for each age group:   - Under 5: $505 \times 0.085 = 42.925$   - 5 to 14: $505 \times 0.104 = 52.52$   - 15 to 64: $505 \times 0.682 = 344.41$   - 65 and older: $505 \times 0.129 = 65.145$2. Calculate $\chi^2$ using the formula:   - Under 5: $\frac{(45 - 42.925)^2}{42.925} = 0.100$   - 5 to 14: $\frac{(50 - 52.52)^2}{52.52} = 0.121$   - 15 to 64: $\frac{(367 - 344.41)^2}{344.41} = 1.804$   - 65 and older: $\frac{(43 - 65.145)^2}{65.145} = 7.206$3. Sum the values: $0.100 + 0.121 + 1.804 + 7.206 = 9.231$Therefore, the chi-square statistic for the sample is 9.231.

Question 16

Suppose the age distribution of the Canadian population and the age distribution of a random sample of 470 residents in the Indian community of Red Lake are shown below. $$\begin{array} { l l l } 
& & \text { Observed Number } \
\text { Age (years) } & \text { Percent of Canadian Population } & \text { in Red Lake Village } \
\hline \text { Under } 5 & 5.9 \% & 44 \
5 \text { to } 14 & 13.7 \% & 58 \
15 \text { to } 64 & 69.9 \% & 324 \
65 \text { and older } & 10.5 \% & 44 \
\hline
\end{array}$$ Use to test the claim that the age distribution of the general Canadian population fits the age distribution of the residents of Red Lake Village. What is the level of significance?
$$\alpha = 0.01$$
A)0.05
B)0.90
C)0.95
D)0.01
E)0.75

Accepted Answer

The level of significance given in the problem is $\alpha = 0.01$, which directly answers the question.

Question 17

A transect is an archaeological study area that is mile wide and 1 mile long. A site in a transect is the location of a significant archaeological find. Let x represent the number of sites per transect. In a section of Chaco Canyon, a large number of transects showed that x has a population variance In a different section of Chaco Canyon, a random sample of 24 transects gave a sample variance for the number of sites per transect. Use an to test the claim that the variance in the new section is greater than 36.7. $\frac { 1 } { 5 }$ $\sigma ^ { 2 } = 36.7$ $s ^ { 2 } = 51.7$ $\alpha = 0.05$ Find the value of the chi-square statistic for the sample.
A)35.22
B)47.63
C)49.61
D)33.81
E)45.64

Accepted Answer

The answer of A transect is an archaeological study area...

Question 18

The following table shows the Myers-Briggs personality preferences for a random sample of 409 people in the listed professions. $$\begin{array} { l | l | l | l } 
\text { Occupation } & \text { Extroverted } & \text { Introverted } & \text { Row Total } \
\hline \text { Clergy (all denominations) } & 62 & 44 & 106 \
\hline \text { M.D. } & 70 & 94 & 164 \
\hline \text { Lawyer } & 54 & 85 & 139 \
\hline \text { Column Total } & 186 & 223 & 409 \
\hline
\end{array}$$ Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.1 level of significance. What are the degrees of freedom?
A)4
B)3
C)2
D)5
E)6

Accepted Answer

The degrees of freedom for a chi-square test of independence are calculated as (number of rows - 1) * (number of columns - 1). Here, we have 3 occupations (rows) and 2 personality preferences (columns), so the degrees of freedom are (3-1)*(2-1) = 2*1 = 2.

Question 19

The following table shows the Myers-Briggs personality preferences for a random sample of 406 people in the listed professions. Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.1 level of significance. Depending on the P-value, will you reject or fail to reject the null hypothesis of independence?
$$\begin{array} { l | l | l | l } 
\text { Occupation } & \text { Extroverted } & \text { Introverted } & \text { Row Total } \
\hline \text { Clergy (all denominations) } & 62 & 44 & 106 \
\hline \text { M.D. } & 66 & 94 & 160 \
\hline \text { Lawyer } & 56 & 84 & 140 \
\hline \text { Column Total } & 184 & 222 & 406 \
\hline
\end{array}$$
A)Since the P-value is greater than $\alpha$, we fail to reject the null hypothesis that the Myers-Briggs personality preference and profession are not independent. At 0.1 level of significance, we conclude that the Myers-Briggs personality preference and profession are independent.
B)Since the P-value is greater than $\alpha$, we reject the null hypothesis that the Myers-Briggs personality preference and profession are not independent. At 0.1 level of significance, we conclude that the Myers-Briggs personality preference and profession are independent.
C)Since the P-value is less than $\alpha$, we reject the null hypothesis that the Myers-Briggs personality preference and profession are not independent. At 0.1 level of significance, we conclude that the Myers-Briggs personality preference and profession are independent.
D)Since the P-value is less than $\alpha$, we reject the null hypothesis that the Myers-Briggs personality preference and profession are independent. At 0.1 level of significance, we conclude that the Myers-Briggs personality preference and profession are not independent.
E)Since the P-value is less than $\alpha$, we fail to reject the null hypothesis that the Myers-Briggs personality preference and profession are independent. At 0.1 level of significance, we conclude that the Myers-Briggs personality preference and profession are not independent.

Accepted Answer

If the P-value is less than the level of significance (0.1 in this case), it indicates that the observed data are unlikely under the null hypothesis of independence. Therefore, we reject the null hypothesis, suggesting that there is a significant association between Myers-Briggs personality preference and profession.

Question 20

The following table shows the Myers-Briggs personality preferences for a random sample of 408 people in the listed professions. $$\begin{array} { l | l | l | l } 
\text { Occupation } & \text { Extroverted } & \text { Introverted } & \text { Row Total } \
\hline \text { Clergy (all denominations) } & 60 & 48 & 108 \
\hline \text { M.D. } & 70 & 91 & 161 \
\hline \text { Lawyer } & 58 & 81 & 139 \
\hline \text { Column Total } & 188 & 220 & 408 \
\hline
\end{array}$$ Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.01 level of significance. If and the degrees of freedom are 2, find (or estimate) the P-value of the sample test statistic.
$$\chi ^ { 2 } = 5.402$$
A)0.05 < P-Value < 0.10
B)P-Value > 0.5
C)P-Value < 0.005
D)0.025 < P-Value < 0.05
E)0.25 < P-Value < 0.5

Accepted Answer

For a chi-square test with 2 degrees of freedom and a test statistic of 5.402, the P-value falls in the range of 0.05 < P-Value < 0.10. This is because the chi-square distribution table or calculator would show that a chi-square value of around 5.402 does not reach the critical value for significance levels below 0.05 but is significant for levels above it, indicating the P-value falls within this range.

Quiz 11: Additional Topics Using Inference