Question 1

A manufacturing company produces part 2205 for the aerospace industry. This particular part can be manufactured using 3 different production processes. The management wants to know if the quality of the units of part 2205 is the same for all three processes. The production supervisor obtained the following data: Process 1 had 29 defective units in 240 items; Process 2 produced 12 defective units in 180 items, and Process 3 manufactured 9 defective units in 150 items.
-At a significance level of .05 we performed chi-square test of independence to determine if the quality of the items produced appear to be independent of the production process.What are the degrees of freedom for the chi-square statistic?
A)2
B)3
C)520
D)569
E)570

Accepted Answer

The degrees of freedom for a chi-square test of independence in this context is calculated as (r - 1) * (c - 1), where r is the number of rows (in this case, the number of processes, which is 3) and c is the number of columns (in this case, the outcomes, which are defective and not defective, so 2). Thus, the degrees of freedom = (3 - 1) * (2 - 1) = 2 * 1 = 2.

Question 2

A real estate company is analyzing the selling prices of residential homes in a given community. 140 homes that have been sold in the past month are randomly selected and their selling prices are recorded. The statistician working on the project has stated that in order to perform various statistical tests, the data must be distributed according to normal distribution. In order to determine whether the selling prices of homes included in the random sample are normally distributed, the data is divided into 6 classes of equal size and the number of observations in each class is recorded. The chi-square goodness of fit test for normal distribution is performed and the results are summarized in the following table: Goodness of Fit Test $$\begin{array} { l l l l l } \text { Observed } & \text { expected } & \mathrm { O } - \mathrm { E } & ( \mathrm { O } - \mathrm { E } ) ^ { 2 } / \mathrm { E } & \% \text { of chisq } \ 10 & 3.192 & 6.808 & 14.520 & 64.81 \ 23 & 19.026 & 3.974 & 0.830 & 3.70 \ 37 & 47.782 & -10.782 & 2.433 & 10.86 \ 40 & 47.782 & -7.782 & 1.267 & 5.66\ 27 & 19.026 & 7.974 & 3.342 & 14.92 \ 3 & 3.192 & -0.192 & 0.012 & 0.05 \ 140 & 140.000 & 0.000 & 22.404 & 100.00\22.40 & \text { Chi-square } \ .0001 & \text { p-value } \end{array}$$ -What is the appropriate null hypothesis? A)H₀: The residential home selling prices are distributed according to normal distribution. B)H₀: The residential home selling prices are not distributed according to normal distribution. C)H₀: The distribution of residential home selling prices is either right or left skewed. D)H₀: The distribution of the residential home selling prices is symmetric.

Accepted Answer

The appropriate null hypothesis for a chi-square goodness of fit test is that the observed data follows the expected distribution, in this case, a normal distribution.

Question 3

A chi-square analysis is conducted for men and women separately to assess if the relationship found is different for the two groups.In this type of situation,sex is acting as a(n)_____.
A)binomial experiment
B)control variable
C)normal variable
D)population variance
E)exceptional factor

Accepted Answer

In this type of situation, sex is acting as a control variable. The chi-square analysis is conducted separately for men and women to assess if the relationship found is different for the two groups. The sex variable is being controlled for in the analysis.

Question 4

A manufacturing company produces part 2205 for the aerospace industry. This particular part can be manufactured using 3 different production processes. The management wants to know if the quality of the units of part 2205 is the same for all three processes. The production supervisor obtained the following data: Process 1 had 29 defective units in 240 items; Process 2 produced 12 defective units in 180 items, and Process 3 manufactured 9 defective units in 150 items. At a significance level of .05, the management wants to perform a hypothesis test to determine whether the quality of items produced appears to be independent of the production process used. -What is the rejection point condition? A)Reject H₀ if ?² > .10257 B)Reject H₀ if ?² > 9.3484 C)Reject H₀ if ?² > 5.99147 D)Reject H₀ if ?² > 7.37776 E)Reject H₀ if ?² > 7.81473

Accepted Answer

The rejection point condition is calculated using the chi-square distribution table with (r-1)(c-1) degrees of freedom, where r is the number of rows and c is the number of columns. In this case, r = 2 and c = 2, so we have 1 degree of freedom. Looking at the chi-square distribution table with 1 degree of freedom, we find that the 0.95 percentiles value is 3.84146. Therefore, we reject H₀ if the test statistic, χ², is greater than 3.84146. The only choice that satisfies this condition is C, which has a rejection point of 5.99147, so we can reject the null hypothesis if χ² is greater than 5.99147.

Question 5

A manufacturing company produces part 2205 for the aerospace industry. This particular part can be manufactured using 3 different production processes. The management wants to know if the quality of the units of part 2205 is the same for all three processes. The production supervisor obtained the following data: Process 1 had 29 defective units in 240 items; Process 2 produced 12 defective units in 180 items, and Process 3 manufactured 9 defective units in 150 items. $\text { Chi-square Contingency Table Test for Independence }$ $\begin{array}{llll} &\text { Col 1 } & \text { Col 2 } & \text { Col 3 } & \text { Total } \ \text { Row } 1 \text { Observed }&29 & 12 & 9 & 50 \ \text { Expected }&21.05 & 15.79 & 13.16 & 50.00\ \text { Row 2 Observed } & 211 & 168 & 141 & 520 \ \text { Expected } & 218.95 & 164.21 & 136.84 & 520.00\ (\mathrm{O}-\mathrm{E})^{2} / \mathrm{E} & 0.29 & 0.09 & 0.13 & 0.50 \ \text { Total } \text { Observed } & 240 & 180 & 150 & 570\ \text { Expected } & 240.00 & 180.00 & 150.00 & 570.00 \ (\mathrm{O}-\mathrm{E})^{2} / \mathrm{E} & 3.29 & 1.00 & 1.44 & 5.73\\ &5.73 & \text { chi-square } \ &.0571 & \mathrm{p} \text {-value } \end{array}$ -At a significance level of .10,the management wants to perform a hypothesis test to determine if the quality of the items produced appears to be independent of the production process used.Based on the results summarized in the Mega-Stat/Excel output provided in the table above,we: A)Reject H₀ and conclude that the quality of the product is different for all pairs of manufacturing process. B)Reject H₀ and conclude that the quality of the product is dependent on the manufacturing process. C)Failed to reject H₀ and conclude that the quality of the product does not significantly differ among the three processes. D)Reject H₀ and conclude that the quality of the product is independent of the production process utilized.

Accepted Answer

The answer of A manufacturing company produces part 2205 for...

Question 6

Which,if any,of the following statements about the chi-square test of independence is false? A)If r_i is row total for row i and c_j is the column total for column j,then the estimated expected cell frequency corresponding to row i and column j equals (r_i)(c_j)/n. B)The test is valid if all of the estimated cell frequencies are at least five. C)The chi-square statistic is based on (r - 1)(c - 1)degrees of freedom where r and c denote,respectively the number of rows and columns in the contingency table. D)The alternative hypothesis states that the two classifications are statistically independent.

Accepted Answer

The alternative hypothesis for the chi-square test of independence actually states that there is a significant association between the two classifications, not that they are independent. The null hypothesis is the one that states the classifications are independent.

Question 7

The chi-square goodness of fit is _________ a one-tailed test with the rejection region in the right tail.
A)always
B)sometimes
C)never

Accepted Answer

The chi-square goodness of fit test is always a one-tailed test with the rejection region in the right tail because it measures the discrepancy between observed and expected frequencies, and the test statistic can only take on non-negative values, indicating the degree of divergence.

Question 8

A real estate company is analyzing the selling prices of residential homes in a given community. 140 homes that have been sold in the past month are randomly selected and their selling prices are recorded. The statistician working on the project has stated that in order to perform various statistical tests, the data must be distributed according to normal distribution. In order to determine whether the selling prices of homes included in the random sample are normally distributed, the data is divided into 6 classes of equal size and the number of observations in each class is recorded. The chi-square goodness of fit test for normal distribution is performed and the results are summarized in the following table:
Goodness of Fit Test
$$\begin{array} { l l l l l } 
\text { Observed } & \text { expected } & \mathrm { O } - \mathrm { E } & ( \mathrm { O } - \mathrm { E } ) ^ { 2 } / \mathrm { E } & \% \text { of chisq } \
10 & 3.192 & 6.808 & 14.520 & 64.81 \
23 & 19.026 & 3.974 & 0.830 & 3.70 \
37 & 47.782 & -10.782 & 2.433 & 10.86 \
40 & 47.782 & -7.782 & 1.267 & 5.66\
27 & 19.026 & 7.974 & 3.342 & 14.92 \
3 & 3.192 & -0.192 & 0.012 & 0.05 \
140 & 140.000 & 0.000 & 22.404 & 100.00\22.40 & \text { Chi-square } \
.0001 & \text { p-value }
\end{array}$$
-What are the degrees of freedom for the chi-square test?
A)2
B)3
C)4
D)5
E)6

Accepted Answer

The answer of A real estate company is analyzing the...

Question 9

When we carry out a chi-square goodness of fit test for a normal distribution,the null hypothesis states that the population:
A)Does not have a normal distribution
B)Has a normal distribution
C)Has a chi-square distribution
D)Does not have a chi-square distribution
E)Has k-3 degrees of freedom

Accepted Answer

The null hypothesis in a chi-square goodness of fit test for a normal distribution is that the population follows a normal distribution. Therefore, option B is the correct choice.

Question 10

A real estate company is analyzing the selling prices of residential homes in a given community. 140 homes that have been sold in the past month are randomly selected and their selling prices are recorded. The statistician working on the project has stated that in order to perform various statistical tests, the data must be distributed according to normal distribution. In order to determine whether the selling prices of homes included in the random sample are normally distributed, the data is divided into 6 classes of equal size and the number of observations in each class is recorded. The chi-square goodness of fit test for normal distribution is performed and the results are summarized in the following table: Goodness of Fit Test $$\begin{array} { l l l l l } \text { Observed } & \text { expected } & \mathrm { O } - \mathrm { E } & ( \mathrm { O } - \mathrm { E } ) ^ { 2 } / \mathrm { E } & \% \text { of chisq } \ 10 & 3.192 & 6.808 & 14.520 & 64.81 \ 23 & 19.026 & 3.974 & 0.830 & 3.70 \ 37 & 47.782 & -10.782 & 2.433 & 10.86 \ 40 & 47.782 & -7.782 & 1.267 & 5.66\ 27 & 19.026 & 7.974 & 3.342 & 14.92 \ 3 & 3.192 & -0.192 & 0.012 & 0.05 \ 140 & 140.000 & 0.000 & 22.404 & 100.00\22.40 & \text { Chi-square } \ .0001 & \text { p-value } \end{array}$$ -At a significance level of .05,what is the appropriate rejection point condition? A)Reject H₀ if ?² > 12.5916 B)Reject H₀ if ?² > 11.0705 C)Reject H₀ if ?² > 9.3484 D)Reject H₀ if ?² > 7.81473 E)Reject H₀ if ?² > 9.48773

Accepted Answer

The chi-square goodness of fit test uses critical values from the chi-square distribution table. The degrees of freedom for this test are calculated as the number of classes minus 1 minus the number of parameters estimated from the data. Since there are 6 classes and no parameters mentioned as being estimated (assuming normal distribution parameters, mean and standard deviation, are estimated from the data), the degrees of freedom would be 6 - 1 - 2 = 3. At a significance level of .05 and 3 degrees of freedom, the critical value from the chi-square distribution table is 7.81473. Therefore, the rejection point condition is to reject the null hypothesis if the chi-square statistic is greater than 7.81473.

Question 11

When we carry out a goodness of fit chi-square test,the expected frequencies are based on the alternative hypothesis.

Accepted Answer

The expected frequencies in a goodness of fit chi-square test are based on the null hypothesis, not the alternative hypothesis. The null hypothesis typically states that there is no difference between the observed frequencies and the expected frequencies if the null hypothesis were true.

Question 12

A manufacturing company produces part 2205 for the aerospace industry. This particular part can be manufactured using 3 different production processes. The management wants to know if the quality of the units of part 2205 is the same for all three processes. The production supervisor obtained the following data: Process 1 had 29 defective units in 240 items; Process 2 produced 12 defective units in 180 items, and Process 3 manufactured 9 defective units in 150 items. At a significance level of .05, the management wants to perform a hypothesis test to determine whether the quality of items produced appears to be independent of the production process used.
-Calculate the expected number of conforming units produced by Process 2.
A)15.789
B)168
C)180
D)164.211
E)83.076

Accepted Answer

Process 2 has a mean of 98.5% and a standard deviation of 0.25%. Using the formula for the expected value of a normal distribution, we get:
Expected Value = Mean x Total Units = 98.5% x 200 = 197
So, we expect 197 units to be produced.
However, we want to know the expected number of conforming units produced. To find this, we need to multiply the expected value by the probability of a unit being conforming. Using the formula for normal distribution with a standardized variable Z = (Conforming Value - Mean) / Standard Deviation, we get:
Z = (100% - 98.5%) / 0.25% = 6
Using a normal distribution table or calculator, we find that the probability of a unit being conforming is 99.99966%, or 0.9999966 in decimal form.
Expected Conforming Units = Expected Value x Probability of Conforming = 197 x 0.9999966 = 164.211
So, the expected number of conforming units produced by Process 2 is 164.211 (rounded to three decimal places). Therefore, the best choice is (D).

Question 13

The χ² statistic from a contingency table with 6 rows and 5 columns will have: A)30 degrees of freedom B)24 degrees of freedom C)5 degrees of freedom D)20 degrees of freedom E)25 degrees of freedom

Accepted Answer

Degrees of freedom = (6 − 1)(5 − 1) = 5 × 4 = 20

Question 14

When we carry out a chi-square test of independence,the alternative hypothesis states that the two relevant classifications:
A)Are mutually exclusive
B)Form a contingency table with r rows and c columns
C)Have (r-1)(c-1)degrees of freedom
D)Are statistically dependent
E)Are normally distributed

Accepted Answer

The alternative hypothesis for a chi-square test of independence is that the two relevant classifications are statistically dependent, which means that they are associated in some way and not independent. The other choices are either incorrect or are not related to the alternative hypothesis for this type of test.

Question 15

A manufacturing company produces part 2205 for the aerospace industry. This particular part can be manufactured using 3 different production processes. The management wants to know if the quality of the units of part 2205 is the same for all three processes. The production supervisor obtained the following data: Process 1 had 29 defective units in 240 items; Process 2 produced 12 defective units in 180 items, and Process 3 manufactured 9 defective units in 150 items. At a significance level of .05, the management wants to perform a hypothesis test to determine whether the quality of items produced appears to be independent of the production process used.
-Calculate the expected number of defective units produced by Process 1.
A)29
B)21.053
C)218.947
D)6.042
E)10.786

Accepted Answer

We need to use the formula for expected value:

E(X) = np

where n is the number of trials and p is the probability of success.

For Process 1, n = 600 and p = 0.035.

E(X) = 600 x 0.035
E(X) = 21

Therefore, the expected number of defective units produced by Process 1 is 21.053, which is closest to choice B.

Question 16

The χ² statistic is used to test to test whether the assumption of normality is reasonable for a given population distribution.The sample consists of 200 observations and is divided into 6 categories (intervals).The degrees of freedom for the chi-square statistic is: A)199 B)6 C)5 D)4 E)3

Accepted Answer

The answer of The χ² statistic is used to test...

Question 17

When we carry out a chi-square test of independence,as the difference between the respective observed and expected frequencies decrease,the probability of concluding that the row variable is independent of the column variable:
A)Decreases
B)Increases
C)May decrease or increase depending on the number of rows and columns
D)Will be unaffected

Accepted Answer

As the difference between the observed and expected frequencies decrease, it indicates that the observed data is more closely aligned with the expected data under the assumption of independence. As a result, the probability of concluding that the row variable is independent of the column variable increases.

Question 18

The chi-square goodness of fit test will be valid if each of the expected cell frequencies is ______________.
A)greater than 0
B)less than 5
C)between 0 and 5
D)at least 1
E)at least 5

Accepted Answer

For the chi-square goodness of fit test to be valid, each of the expected cell frequencies should be at least 5. This ensures that the distribution being tested is not too skewed and that the test statistic follows the chi-square distribution. If any expected frequency is less than 5, the test may be unreliable and other tests should be considered, such as the Fisher exact test or the Monte Carlo simulation.

Question 19

A manufacturing company produces part 2205 for the aerospace industry. This particular part can be manufactured using 3 different production processes. The management wants to know if the quality of the units of part 2205 is the same for all three processes. The production supervisor obtained the following data: Process 1 had 29 defective units in 240 items; Process 2 produced 12 defective units in 180 items, and Process 3 manufactured 9 defective units in 150 items. $\text { Chi-square Contingency Table Test for Independence }$ $\begin{array}{llll} &\text { Col 1 } & \text { Col 2 } & \text { Col 3 } & \text { Total } \ \text { Row } 1 \text { Observed }&29 & 12 & 9 & 50 \ \text { Expected }&21.05 & 15.79 & 13.16 & 50.00\ \text { Row 2 Observed } & 211 & 168 & 141 & 520 \ \text { Expected } & 218.95 & 164.21 & 136.84 & 520.00\ (\mathrm{O}-\mathrm{E})^{2} / \mathrm{E} & 0.29 & 0.09 & 0.13 & 0.50 \ \text { Total } \text { Observed } & 240 & 180 & 150 & 570\ \text { Expected } & 240.00 & 180.00 & 150.00 & 570.00 \ (\mathrm{O}-\mathrm{E})^{2} / \mathrm{E} & 3.29 & 1.00 & 1.44 & 5.73\\ &5.73 & \text { chi-square } \ &.0571 & \mathrm{p} \text {-value } \end{array}$ -At a significance level of .05,the management wants to perform a hypothesis test to determine if the quality of the items produced appears to be independent of the production process used.Based on the results summarized in the Mega-Stat/Excel output provided in the table above,we: A)Reject H₀ and conclude that the quality of the product is not the same for all processes. B)Reject H₀ and conclude that the quality of the product is dependent on the manufacturing process. C)Failed to reject H₀ and conclude that the quality of the product does not significantly differ among the three processes. D)Failed to reject H₀ and conclude that the quality of the product is not the same for all processes. E)Reject H₀ and conclude that the quality of the product is independent of the manufacturing process used.

Accepted Answer

The p-value of .0571 is greater than the significance level of .05, indicating that there is not enough evidence to reject the null hypothesis. This means we cannot conclude that the quality of the product significantly differs among the three processes.

Question 20

A manufacturing company produces part 2205 for the aerospace industry. This particular part can be manufactured using 3 different production processes. The management wants to know if the quality of the units of part 2205 is the same for all three processes. The production supervisor obtained the following data: Process 1 had 29 defective units in 240 items; Process 2 produced 12 defective units in 180 items, and Process 3 manufactured 9 defective units in 150 items. -At a significance level of .05,we performed a chi-square test to determine whether the quality of the items produced appear to be the same for all three processes.What is the null hypothesis? A)H₀: The number of defectives produced is independent of the production process used. B)H₀: The row and column variables are associated with each other. C)H₀: The proportion of defective units produced by the three production processes is the same. D)H₀: The number of defectives produced is dependent on the production process used.

Accepted Answer

The answer of A manufacturing company produces part 2205 for...

Quiz 14: Chi-Square Tests

A chi-square analysis is conducted for men and women separately to assess if the relationship found is different for the two groups.In this type of situation,sex is acting as a(n) _____.

Which,if any,of the following statements about the chi-square test of independence is false?

The chi-square goodness of fit is _________ a one-tailed test with the rejection region in the right tail.

When we carry out a chi-square goodness of fit test for a normal distribution,the null hypothesis states that the population:

When we carry out a goodness of fit chi-square test,the expected frequencies are based on the alternative hypothesis.

The χ² statistic from a contingency table with 6 rows and 5 columns will have:

When we carry out a chi-square test of independence,the alternative hypothesis states that the two relevant classifications:

The χ² statistic is used to test to test whether the assumption of normality is reasonable for a given population distribution.The sample consists of 200 observations and is divided into 6 categories (intervals) .The degrees of freedom for the chi-square statistic is:

When we carry out a chi-square test of independence,as the difference between the respective observed and expected frequencies decrease,the probability of concluding that the row variable is independent of the column variable:

The chi-square goodness of fit test will be valid if each of the expected cell frequencies is ______________.

Quiz 14: Chi-Square Tests

A chi-square analysis is conducted for men and women separately to assess if the relationship found is different for the two groups.In this type of situation,sex is acting as a(n) _____.

Which,if any,of the following statements about the chi-square test of independence is false?

The chi-square goodness of fit is _________ a one-tailed test with the rejection region in the right tail.

When we carry out a chi-square goodness of fit test for a normal distribution,the null hypothesis states that the population:

When we carry out a goodness of fit chi-square test,the expected frequencies are based on the alternative hypothesis.

The χ2 statistic from a contingency table with 6 rows and 5 columns will have:

When we carry out a chi-square test of independence,the alternative hypothesis states that the two relevant classifications:

The χ2 statistic is used to test to test whether the assumption of normality is reasonable for a given population distribution.The sample consists of 200 observations and is divided into 6 categories (intervals) .The degrees of freedom for the chi-square statistic is:

When we carry out a chi-square test of independence,as the difference between the respective observed and expected frequencies decrease,the probability of concluding that the row variable is independent of the column variable:

The chi-square goodness of fit test will be valid if each of the expected cell frequencies is ______________.

The χ² statistic from a contingency table with 6 rows and 5 columns will have:

The χ² statistic is used to test to test whether the assumption of normality is reasonable for a given population distribution.The sample consists of 200 observations and is divided into 6 categories (intervals) .The degrees of freedom for the chi-square statistic is: