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Q:
Consider a set of 50 measurements with mean 50.2 and standard deviation 18.7 and with the following observed frequencies.It is desired to test whether these measurements came from a normal population. How many degrees of freedom are associated with the chi-square test?
Q:
Consider a set of 50 measurements with mean 50.2 and standard deviation 18.7 and with the following observed frequencies.It is desired to test whether these measurements came from a normal population.Calculate the expected frequency for the interval 80 and higher.
Q:
Consider a set of 50 measurements with mean 50.2 and standard deviation 18.7 and with the following observed frequencies.It is desired to test whether these measurements came from a normal population. Calculate the expected frequency for the interval 60-79.99.
Q:
Consider a set of 50 measurements with mean 50.2 and standard deviation 18.7 and with the following observed frequencies.It is desired to test whether these measurements came from a normal population. Calculate the expected frequency for the interval 40-59.99.
Q:
Consider a set of 50 measurements with mean 50.2 and standard deviation 18.7 and with the following observed frequencies.It is desired to test whether these measurements came from a normal population. Calculate the expected frequency for the interval 0-39.99.
Q:
Consider the 3 x 2 contingency table below. At a significance level of .05, test H0: the factors A and B are independent.
Q:
Consider the 3 x 2 contingency table below.At = .05, determine the tabular value of the chi-square statistic used to test for the independence of Factors A and B.
Q:
Consider the 3 x 2 contingency table below. How many degrees of freedom are associated with the chi-square test?
Q:
Consider the 3 x 2 contingency table below. Compute the expected frequencies in row 3.
Q:
Consider the 3 x 2 contingency table below. Compute the expected frequencies in row 2.
Q:
Consider the 3 x 2 contingency table below. Compute the expected frequencies in row 1.
Q:
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 a normal distribution. In order to determine whether the selling prices of homes included in the random sample are normally distributed, the statistician divides the data into 6 classes of equal size and records the number of observations in each class. She then performs a chi-square goodness-of-fit test for normal distribution. The results are summarized in the following table. At a significance level of .05, we
A. reject H0; conclude that the residential home selling prices are not distributed according to a normal distribution.
B. do not reject H0; conclude that the residential home selling prices are not distributed according to a normal distribution.
C. reject H0; conclude that the residential home selling prices are distributed according to a normal distribution.
D. do not reject H0; conclude that the residential home selling prices are distributed according to a normal distribution.
Q:
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 a normal distribution. In order to determine whether the selling prices of homes included in the random sample are normally distributed, the statistician divides the data into 6 classes of equal size and records the number of observations in each class. She then performs a chi-square goodness-of-fit test for normal distribution. The results are summarized in the following table.At a significance level of .05, what is the appropriate rejection point condition?A. Reject H0 if x2 > 12.5916B. Reject H0 if x2 > 11.0705C. Reject H0 if x2 > 9.3484D. Reject H0 if x2 > 7.81473E. Reject H0 if x2 > 9.48773
Q:
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 a normal distribution. In order to determine whether the selling prices of homes included in the random sample are normally distributed, the statistician divides the data into 6 classes of equal size and records the number of observations in each class. She then performs a chi-square goodness-of-fit test for normal distribution. The results are summarized in the following table What are the degrees of freedom for the chi-square test?
A. 2
B. 3
C. 4
D. 5
E. 6
Q:
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 a normal distribution. In order to determine whether the selling prices of homes included in the random sample are normally distributed, the statistician divides the data into 6 classes of equal size and records the number of observations in each class. She then performs a chi-square goodness-of-fit test for normal distribution. The results are summarized in the following table. What is the appropriate null hypothesis?
A. H0: The residential home selling prices are distributed according to a normal distribution.
B. H0: The residential home selling prices are not distributed according to a 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.
E. None of the other answers is correct.
Q:
A manufacturing company produces part QV2Y 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 QV2Y 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.
Chi-Square Contingency Table Test for Independence 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 MegaStat/Excel output provided in the table above, we
A. do not 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. do not 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.
Q:
A manufacturing company produces part QV2Y 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 QV2Y 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.
Chi-Square Contingency Table Test for Independence 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 MegaStat/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. do not reject H0, and conclude that the quality of the product does not significantly differ among the three processes.
D. do not 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.
Q:
A manufacturing company produces part QV2Y 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 QV2Y 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 x2 > .10257B. Reject H0 if x2 > 9.3484C. Reject H0 if x2 > 5.99147D. Reject H0 if x2 > 7.37776E. Reject H0 if x2 > 7.81473
Q:
A manufacturing company produces part QV2Y 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 QV2Y 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 of independence to determine if the quality of the items produced appears to be independent of the production process. What are the degrees of freedom for the chi-square statistic?
A. 2
B. 3
C. 50
D. 520
E. 570
Q:
A special version of the chi-square goodness-of-fit test that involves testing the null hypothesis that all of the multinomial probabilities are equal is called the test for ___________.
A. goodness of fit
B. statistical independence
C. normality
D. homogeneity
Q:
In performing a chi-square goodness-of-fit test with multinomial probabilities, the ___________ the difference between observed and expected frequencies, the higher the probability of concluding that the probabilities specified in the null hypothesis are correct.A. largerB. smaller
Q:
In performing a chi-square test of independence, as the differences between respective observed and expected frequencies _________, the probability of concluding that the row variable is independent of the column variable increases.
A. stay the same
B. decrease
C. increase
D. double
Q:
As the difference between observed frequency and expected frequency _______________, the probability of rejecting the null hypothesis increases.
A. stays the same
B. decreases
C. increases
D. goes to 0
Q:
In performing a chi-square goodness-of-fit test for a normal distribution, a researcher wants to make sure that all of the expected cell frequencies are at least five. The sample is divided into 7 intervals. The second through the sixth intervals all have expected cell frequencies of at least five. The first and the last intervals have expected cell frequencies of 1.5 each. After adjusting the number of intervals, the degrees of freedom for the chi-square statistic is ____.
A. 2
B. 3
C. 5
D. 7
Q:
While a binomial distribution describes count data that can be classified into one of two mutually exclusive categories, a __________________ distribution describes count data that are classified into more than two mutually exclusive categories.
A. normal
B. skewed
C. uniform
D. multinomial
Q:
The chi-square goodness-of-fit test for multinomial probabilities with 5 categories has _____ degrees of freedom.
A. 5
B. 4
C. 3
D. 6
Q:
An experiment consists of 400 observations and four mutually exclusive groups. If the probability of a randomly selected item being classified into any of the four groups is equal, then the expected number of items that will be classified into group 1 is _____.
A. 25
B. 100
C. 125
D. 150
Q:
The number of degrees of freedom associated with a chi-square test for independence based upon a contingency table with 4 rows and 3 columns is _____.
A. 7
B. 12
C. 5
D. 6
Q:
In performing a chi-square goodness-of-fit test for a normal distribution, if there are 7 intervals, then the number of degrees of freedom for the chi-square statistic is ______.
A. 7
B. 3
C. 4
D. 6
Q:
A manufacturing company produces part QV2Y 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 QV2Y 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 appears 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. Both -H0: The number of defectives produced is independent of the production process used." and -H0: The proportion of defective units produced by the three production processes is the same." are correct or at least acceptable ways of stating the null hypothesis.E. All of the other choices are acceptable ways of stating the null hypothesis.
Q:
When we carry out a chi-square test of independence, as the differences 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.
Q:
Which, if any, of the following statements about the chi-square test of independence is false?A. If ri is the row total for row i and cj is the column total for column j, then the estimated expected cell frequency corresponding to row i and column j equals (ri)(cj)/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.E. All of the other statements about the chi-square test of independence are true.
Q:
When we carry out a chi-square test of independence, the alternate hypothesis states that the two relevant classificationsA. 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.
Q:
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.
Q:
The chi-square goodness-of-fit test will be valid if the average 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
Q:
The χ2 statistic is used to test whether the assumption of normality is reasonable for a given population distribution. The sample consists of 5,000 observations and is divided into 6 categories (intervals). The degrees of freedom for the chi-square statistic are
A. 4,999.
B. 6.
C. 5.
D. 4.
E. 3.
Q:
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
Q:
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.
Q:
When we carry out a goodness-of-fit chi-square test, the expected frequencies are based on the alternative hypothesis.
Q:
When we carry out a chi-square test of independence, the expected frequencies are based on the null hypothesis.
Q:
When we carry out a chi-square test of independence, in the alternative hypothesis we state that the two classifications are statistically independent.
Q:
When we carry out a chi-square test of independence, the chi-square statistic is based on (r x c) - 1 degrees of freedom, where r and c denote, respectively, the number of rows and columns in the contingency table.
Q:
When we carry out a chi-square test of independence, if ri is the row total for row i and cj is the column total for column j, then the estimated expected cell frequency corresponding to row i and column j equals (ri)(cj)/n.
Q:
In performing a chi-square goodness-of-fit test with multinomial probabilities, the smaller the difference between observed and expected frequencies, the higher the probability of concluding that the probabilities specified in the null hypothesis are correct.
Q:
In performing a chi-square test of independence, as the difference between the respective observed and expected frequencies calculated by assuming independence decreases, the probability of concluding that the row variable is independent of the column variable decreases.
Q:
The chi-square goodness-of-fit test can only be used to test whether a population has specified multinomial probabilities or to test if a sample has been selected from a normally distributed population. It cannot be used if sample data come from other distribution forms, such as the Poisson.
Q:
When using the chi-square goodness-of-fit test, if the value of the chi-square statistic is large enough, we reject the null hypothesis.
Q:
A fastener manufacturing company uses a chi-square goodness-of-fit test to determine if a population of all lengths of -inch bolts it manufactures is distributed according to a normal distribution. If we reject the null hypothesis, it is reasonable to assume that the population distribution is at least approximately normally distributed.
Q:
When using a chi-square goodness-of-fit test with multinomial probabilities, the rejection of the null hypothesis indicates that at least one of the multinomial probabilities is not equal to the value stated in the null hypothesis.
Q:
In a contingency table, if all of the expected frequencies equal the observed frequencies, then we can conclude that there is a perfect association between rows and columns.
Q:
In a contingency table, when all the expected frequencies equal the observed frequencies, the calculated χ2 statistic equals zero.
Q:
One use of the chi-square goodness-of-fit test is to determine if specified multinomial probabilities in the null hypothesis are correct.
Q:
The trials of a multinomial probability are assumed to be dependent.
Q:
A multinomial probability distribution describes data that are classified into two or more categories when a multinomial experiment is carried out.
Q:
Expected cell frequencies for a multinomial distribution are calculated by assuming statistical dependence.
Q:
The chi-square distribution is a continuous probability distribution that is skewed to the left.
Q:
The χ2 goodness-of-fit test requires the nominative level of data.
Q:
The actual counts in the cells of a contingency table are referred to as the expected cell frequencies.
Q:
A contingency table summarizes data that has been classified on two dimensions or scales.
Q:
Consider the following partial analysis of variance table from a randomized block design with 10 blocks and 6 treatments.Source SSTreatments 2,477.53Blocks 3,180.48Error 11,661.38Total Determine the degrees of freedom for the blocks.
Q:
Consider the following partial analysis of variance table from a randomized block design with 10 blocks and 6 treatments.Source SSTreatments 2,477.53Blocks 3,180.48Error 11,661.38Total Determine the degrees of freedom for treatments.
Q:
A company that fills one-gallon containers of water has four machines. The quality control manager needs to determine whether the average fill for these machines is the same. For a sample of 19 one-gallon containers, we have the following data of fill measures (x) in quarts. Machine 1 Machine 2 Machine 3 Machine 4N 4 6 5 4 4.03 4.0017 3.974 4.005S 0.0183 0.0117 0.0182 0.0129And the following partial ANOVA table. Source SS DF MS FTreatment 0.002359 Error Total 0.010579
Q:
A company that fills one-gallon containers of water has four machines. The quality control manager needs to determine whether the average fill for these machines is the same. For a sample of 19 one-gallon containers, we have the following data of fill measures (x) in quarts. Machine 1 Machine 2 Machine 3 Machine 4N 4 6 5 4 4.03 4.0017 3.974 4.005S 0.0183 0.0117 0.0182 0.0129And the following partial ANOVA table. Source SS DF MS FTreatment 0.002359 Error Total 0.010579 Determine the degrees of freedom for the treatment, error, and total, and state the critical value of the F statistic at = .05.
Q:
A company that fills one-gallon containers of water has four machines. The quality control manager needs to determine whether the average fill for these machines is the same. For a sample of 19 one-gallon containers, we have the following data of fill measures (x) in quarts. Machine 1 Machine 2 Machine 3 Machine 4N 4 6 5 4 4.03 4.0017 3.974 4.005S 0.0183 0.0117 0.0182 0.0129And the following partial ANOVA table. Source SS DF MS FTreatment 0.002359 Error Total 0.010579 Complete the ANOVA table and calculate F.
Q:
Suppose you are a researcher investigating the annual sales differences among five categories of businesses. The study looks at 55 companies equally divided among categories A, B, C, D, and E.Source SS DF MS FTreatment 583.39 4 145.85 7.501Error 972.18 50 19.4436 Total 1555.57 54 Is there a significant difference in the annual sales of the five business categories at α = .05? Do you reject H0 ?
Q:
Suppose you are a researcher investigating the annual sales differences among five categories of businesses. The study looks at 55 companies equally divided among categories A, B, C, D, and E.
Determine degrees of freedom treatment, degrees of freedom error and degrees of freedom total and state the critical value of the F statistic at α = .05.
Q:
Suppose you are a researcher investigating the annual sales differences among five categories of businesses. The study looks at 55 companies equally divided among categories A, B, C, D, and E.Complete the following ANOVA table and determine the value of the F statistic.Source SS DF MS FTreatment 583.39 Error Total 1555.57
Q:
A researcher has used a one-way analysis of variance model to test whether the average starting salaries differ among recent graduates from the nursing, engineering, business, and education disciplines. She has randomly selected four graduates from each of the four areas.If MSE = 4, and SSTO = 120, complete the following ANOVA table. Is there a significant difference in the starting salaries among the four disciplines? (H0: No difference)Source SS DF MS FTreatment 4 Error Total 120
Q:
A researcher has used a one-way analysis of variance model to test whether the average starting salaries differ among recent graduates from the nursing, engineering, business, and education disciplines. She has randomly selected four graduates from each of the four areas.Determine the degrees of freedom treatment, degrees of freedom error, and degrees of freedom total, and state the critical value of the F statistic at = .05
Q:
A researcher has used a one-way analysis of variance model to test whether the average starting salaries differ among recent graduates from the nursing, engineering, business, and education disciplines. She has randomly selected four graduates from each of the four areas.If MSE = 4, and SSTO = 120, complete the following ANOVA table and determine the value of the F statistic.Source SS DF MS FTreatment Error Total
Q:
Consider the following partial analysis of variance table from a randomized block design with 6 blocks and 4 treatments.Source Sum of SquaresTreatments 15.93Blocks 42.09Error 23.84Total 81.86
Q:
Consider the following partial analysis of variance table from a randomized block design with 6 blocks and 4 treatments.Source Sum of SquaresTreatments 15.93Blocks 42.09Error 23.84Total 81.86
Q:
Consider the following partial analysis of variance table from a randomized block design with 6 blocks and 4 treatments.Source Sum of SquaresTreatments 15.93Blocks 42.09Error 23.84Total 81.86What is the calculated F statistic for blocks?
Q:
Consider the following partial analysis of variance table from a randomized block design with 6 blocks and 4 treatments.Source Sum of SquaresTreatments 15.93Blocks 42.09Error 23.84Total 81.86What is the calculated F statistic for treatments?
Q:
Consider the following partial analysis of variance table from a randomized block design with 6 blocks and 4 treatments.Source Sum of SquaresTreatments 15.93Blocks 42.09Error 23.84Total 81.86What is the mean square error?
Q:
Consider the following partial analysis of variance table from a randomized block design with 6 blocks and 4 treatments.Source Sum of SquaresTreatments 15.93Blocks 42.09Error 23.84Total 81.86What is the block mean square?
Q:
Consider the following partial analysis of variance table from a randomized block design with 6 blocks and 4 treatments.Source Sum of SquaresTreatments 15.93Blocks 42.09Error 23.84Total 81.86What is the treatment mean square?
Q:
Consider the following partial analysis of variance table from a randomized block design with 6 blocks and 4 treatments.Source Sum of SquaresTreatments 15.93Blocks 42.09Error 23.84Total 81.86Determine the degrees of freedom for error.
Q:
Consider the following partial analysis of variance table from a randomized block design with 6 blocks and 4 treatments.Source Sum of SquaresTreatments 15.93Blocks 42.09Error 23.84Total 81.86Calculate the degrees of freedom for blocks.
Q:
Consider the following partial analysis of variance table from a randomized block design with 6 blocks and 4 treatments.Source Sum of SquaresTreatments 15.93Blocks 42.09Error 23.84Total 81.86Determine the degrees of freedom for treatments.