Q&A Hero

Q: The sampling distribution for a goodness-of-fit test is the Poisson distribution.

Q: If one or more parameters are left unspecified in a goodness-of-fit test, they must be estimated from the sample data and one degree of freedom is lost for each parameter that must be estimated.

Q: By combining cells we guard against having an inflated test statistic that could have led us to incorrectly reject the H0.

Q: The goodness-of-fit test is essentially determining if the test statistic is significantly larger than zero.

Q: If the calculated chi-square statistic is large, this is evidence to suggest the fit of the actual data to the hypothesized distribution is not good, and H0should be rejected.

Q: A goodness-of-fit test can decide whether a set of data comes from a specific hypothesized distribution.

Q: A lube and oil change business believes that the number of cars that arrive for service is the same each day of the week. If the business is open six days a week (Monday - Saturday) and a random sample of n = 200 customers is selected, the critical value for testing the hypothesis using a goodness-of-fit test is x2= 9.2363 if the alpha level for the test is set at .10.

Q: If any of the observed frequencies are smaller than 5, then categories should be combined until all observed frequencies are at least 5.

Q: If the sample size is large, the standard normal distribution can be used in place of the chi-square in a goodness-of-fit test for testing whether the population is normally distributed.

Q: It is believed that the number of drivers who are ticketed for speeding on a particular stretch of highway is a Poisson distribution with a mean of 3.5 per hour. A random sample of 100 hours is selected with the following results: X Frequency 0 5 1 10 2 20 3 18 4 20 5 15 6 4 7 6 8 1 9 2 100 Given this information, it can be seen that the cells will need to be combined since the actual number of occurrences at some levels of x is less than 5.

Q: A business with 5 copy machines keeps track of how many copy machines need service on a given day. It believes this is binomially distributed with a probability of p = 0.2 of each machine needing service on any given day. It has collected the following based on a random sample of 100 days.XFrequency028138222374 or 55Total100 Given this information, assuming that all expected values are sufficiently large to use the classes as shown above, the critical value based on a 0.05 level of significance is 9.4877.Answer: TRUE

Q: The Conrad Real Estate Company recently conducted a statistical test to determine whether the number of days that homes are on the market prior to selling is normally distributed with a mean equal to 50 days and a standard deviation equal to 10 days. The sample of 200 homes was divided into 8 groups to form a grouped data frequency distribution. If a chi-square goodness-of-fit test is to be conducted using an alpha = .05, the critical value is 14.0671.

Q: In a goodness-of-fit test, when the null hypothesis is true, the expected value for the chi-square test statistic is zero.

Q: The reason that a decision maker might want to combine groups before performing a goodness-of-fit test is to avoid accepting the null hypothesis due to an inflated value of the test statistic.

Q: When the expected cell frequencies are smaller than 5, the cells should be combined in a meaningful way such that the expected cell frequencies do exceed 5.

Q: The goodness-of-fit test is always a one-tail test with the rejection region in the upper tail.

Q: The logic behind the chi-square goodness-of-fit test is based on determining how far the actual observed frequencies are from the expected frequencies.

Q: If the test statistic for a chi-square goodness-of-fit test is larger than the critical value, the null hypothesis should be rejected.

Q: A goodness-of-fit test can be used to determine whether a set of sample data comes from a specific hypothesized population distribution.

Q: Explain why, in performing a goodness-of-fit test, it is sometimes necessary to combine categories.

Q: The billing department of a national cable service company is conducting a study of how customers pay their monthly cable bills. The cable company accepts payment in one of four ways: in person at a local office, by mail, by credit card, or by electronic funds transfer from a bank account. The cable company randomly sampled 400 customers to determine if there is a relationship between the customer's age and the payment method used. The following sample results were obtained:Based on the sample data, can the cable company conclude that there is a relationship between the age of the customer and the payment method used? Conduct the appropriate test at the alpha= 0.01 level of significance.A) Because x2= 42.2412 > 21.666, do not reject the null hypothesis. Based on the sample data conclude that age and type of payment are independent.B) Because x2= 42.2412 > 21.666, reject the null hypothesis. Based on the sample data conclude that age and type of payment are not independentC) Because x2= 50.3115 > 21.666, do not reject the null hypothesis. Based on the sample data conclude that age and type of payment are independent.D) Because x2= 50.3115 > 21.666, reject the null hypothesis. Based on the sample data conclude that age and type of payment are not independent.

Q: We want to test whether type of car owned (domestic or foreign) is independent of gender. A contingence table is obtained from a sample of 990 people asAt alpha = 0.05 level, we conclude that:A) x2= 3.34 and type of car owned is independent of gender.B) x2= 3.34 and type of car owned is dependent of gender.C) x2= 3.84 and type of car owned is independent of gender.D) x2= 3.84 and type of car owned is dependent of gender.

Q: The degrees of freedom for a contingency table with 11 rows and 10 columns is: A) 11 B) 10 C) 110 D) 90

Q: Many companies use well-known celebrities as spokespeople in their TV advertisements. A study was conducted to determine whether brand awareness of female TV viewers and the gender of the spokesperson are independent. Each in a sample of 300 female TV viewers was asked to identify a product advertised by a celebrity spokesperson. The gender of the spokesperson and whether or not the viewer could identify the product was recorded. The numbers in each category are given below. Male Celebrity Female Celebrity Identified product 41 61 Could not identify 109 89 Referring to these sample data, if the appropriate null hypothesis is tested using a significance level equal to .05, which of the following conclusions should be reached? A) There is a relationship between gender of the celebrity and product identification. B) There is no relationship between gender of the celebrity and product identification. C) The mean number of products identified for males is different than the mean number for females. D) Females have higher brand awareness than males.

Q: A study published in the American Journal of Public Health was conducted to determine whether the use of seat belts in motor vehicles depends on ethnic status in San Diego County. A sample of 792 children treated for injuries sustained from motor vehicle accidents was obtained, and each child was classified according to (1) ethnic status (Hispanic or non-Hispanic) and (2) seat belt usage (worn or not worn) during the accident. The number of children in each category is given in the table below. Hispanic Non-Hispanic Seat belts worn 31 148 Seat belts not worn 283 330 Referring to these data, which of the following conclusions should be reached if the appropriate hypothesis is conducted using an alpha = .05 level? A) The mean value for Hispanics is the same as for Non-Hispanics. B) There is no relationship between whether someone is Hispanic and whether they wear a seat belt. C) The use of seat belts and whether a person is Hispanic or not is statistically related. D) None of the above

Q: In testing a hypothesis that two categorical variables are independent using the x2test, the expected cell frequencies are based on assuming: A) the null hypothesis. B) the alternative hypothesis. C) the normal distribution. D) the variable are related.

Q: When testing for independence in a contingency table with 3 rows and 4 columns, there are ________ degrees of freedom. A) 5 B) 6 C) 7 D) 12

Q: For a chi-square test involving a contingency table, suppose H0is rejected. We conclude that the two variables are: A) curvilinear. B) linear. C) related. D) not related.

Q: How can the degrees of freedom be found in a contingency table with cross-classified data? A) When df are equal to rows minus columns B) When df are equal to rows multiplied by columns C) When df are equal to rows minus 1 multiplied by columns minus 1 D) Total number of cell minus 1

Q: In performing chi-square contingency analysis, to overcome a small expected cell frequency problem, we: A) combine the categories of the row and/or column variables. B) increase the sample size. C) Both A and B D) None of the above

Q: In a chi-square contingency analysis, when expected cell frequencies drop below 5, the calculated chi-square value tends to be inflated and may inflate the true probability of ________ beyond the stated significance level. A) committing a Type I error B) committing a Type II error C) Both A and B D) All of the above

Q: Explain what is meant by interaction between two factors in a two-way analysis of variance study.

Q: Assume you are conducting a two-way ANOVA and have found a significant interaction between the two factors. Explain what conclusions you can make from the results and what if any further steps should be taken.

Q: Under what circumstances would you use a randomized complete block analysis of variance design instead of a one-way analysis of variance?

Q: Explain what is meant by partitioning the sum of squares in a one-way analysis of variance application.

Q: What are the assumptions for a one-way analysis of variance design?

Q: What is meant by the term balanced design in an analysis of variance application?

Q: Examine the following two-factor analysis of variance table:Does the ANOVA table indicate that the levels of factor B have equal means? Use a significance level of 0.05.A) Fail to reject H0. Conclude that there is not sufficient evidence to indicate that at least two levels of Factor B have different mean responses.B) Reject H0. Conclude that there is sufficient evidence to indicate that at least two levels of Factor B have different mean responses.C) Fail to reject H0. Conclude that there is sufficient evidence to indicate that at least two levels of Factor B have different mean responses.D) Reject H0. Conclude that there is not sufficient evidence to indicate that at least two levels of Factor B have different mean responses.

Q: Examine the following two-factor analysis of variance table:Determine if the levels of factor A have equal means. Use a significance level of 0.05.A) Fail to reject H0. Conclude that there is sufficient evidence to indicate that at least two levels of Factor A have different mean responses.B) Fail to reject H0. Conclude that there is not sufficient evidence to indicate that at least two levels of Factor A have different mean responses.C) Reject H0. Conclude that there is sufficient evidence to indicate that at least two levels of Factor A have different mean responses.D) Reject H0. Conclude that there is not sufficient evidence to indicate that at least two levels of Factor A have different mean responses.

Q: Examine the following two-factor analysis of variance table:Determine if interaction exists between factor A and factor B. Use alpha = 0.05.A) Fail to reject H0. Conclude that there is not sufficient evidence to indicate interaction exists between Factor A and Factor BB) Reject H0. Conclude that there is sufficient evidence to indicate interaction exists between Factor A and Factor BC) Fail to reject H0. Conclude that there is sufficient evidence to indicate interaction exists between Factor A and Factor BD) Reject H0. Conclude that there is not sufficient evidence to indicate interaction exists between Factor A and Factor B

Q: Examine the following two-factor analysis of variance table:Complete the analysis of variance table.A) MSA = 40.928, F Factor A =3.35, SSB = 85.35, Factor B df = 3, F Factor B = 2.316, MSAB = 21.859, F Factor AB = 1.8, SSE = 789.29, SSE df = 66, MSE = 12.143B) MSA = 40.928, F Factor A = 3.35, SSB = 85.35, Factor B df = 4, F Factor B = 2.316, MSAB = 21.859, F Factor AB = 2.1 SSE = 789.29, SSE df = 66, MSE = 12.143C) MSA = 40.698, F Factor A = 3.35, SSB = 84.35, Factor B df = 5, F Factor B = 2.316, MSAB = 21.859, F Factor AB = 2.1, SSE = 789.29, SSE df = 65, MSE = 12.143D) MSA = 40.698, F Factor A = 3.35, SSB = 84.35, Factor B df = 3, F Factor B = 2.316, MSAB = 21.859, F Factor AB = 1.8, SSE = 789.29, SSE df = 65, MSE = 12.143

Q: When the world's largest retailer, Walmart, decided to enter the grocery marketplace in a big way with its "Super Stores," it changed the retail grocery landscape in a major way. The other major chains such as Albertsons have struggled to stay competitive. In addition, regional discounters such as WINCO in the western United States have made it difficult for the traditional grocery chains. Recently, a study was conducted in which a "market basket" of products was selected at random from those items offered in three stores in Boise, Idaho: Walmart, Winco, and Albertsons. At issue was whether the mean prices at the three stores are equal or whether there is a difference in prices. The sample data are in the data file called Food Price Comparisons. Using an alpha level equal to 0.05, test to determine whether the three stores have equal population mean prices. If you conclude that there are differences in the mean prices, perform the appropriate posttest to determine which stores have different means. A) There is no difference between the three mean prices. B) Based on the sample data, we conclude that Winco is significantly different (higher) than Albertsons and Walmart in terms of average prices. However, we can make no conclusion about Albertsons and Walmart. C) Based on the sample data, we conclude that Walmart is significantly different (higher) than Albertsons and Winco in terms of average prices. However, we can make no conclusion about Albertsons and Winco. D) Based on the sample data, we conclude that Albertsons is significantly different (higher) than Walmart and Winco in terms of average prices. However, we can make no conclusion about Walmart and Winco.

Q: In a local community there are three grocery chain stores. The three have been carrying out a spirited advertising campaign in which each claims to have the lowest prices. A local news station recently sent a reporter to the three stores to check prices on several items. She found that for certain items each store had the lowest price. This survey didn't really answer the question for consumers. Thus, the station set up a test in which 20 shoppers were given different lists of grocery items and were sent to each of the three chain stores. The sales receipts from each of the three stores are recorded in the data file Groceries. Based on the sample data, which store has the highest average prices? Use Fisher's LSD test if appropriate. A) Store 1 B) Store 2 C) Store 3 D) There is no difference between the average prices.

Q: Applebee's International, Inc., is a U.S. company that develops, franchises, and operates the Applebee's Neighborhood Grill and Bar restaurant chain. It is the largest chain of casual dining restaurants in the country, with over 1,500 restaurants across the United States. The headquarters is located in Overland Park, Kansas. The company is interested in determining if mean weekly revenue differs among three restaurants in a particular city. The file entitled Applebees contains revenue data for a sample of weeks for each of the three locations. If you did conclude that there was a difference in the average revenue, use Fisher's LSD approach to determine which restaurant has the lowest mean sales. A) There is no difference between the average revenues. B) Restaurant 1 has the highest average revenue while there is no evidence of a difference between Restaurant 2's and 3's average revenues. C) Restaurant 3 has the highest average revenue while there is no evidence of a difference between Restaurant 1's and 2's average revenues. D) Restaurant 2 has the highest average revenue while there is no evidence of a difference between Restaurant 1's and 3's average revenues.

Q: Frasier and Company manufactures four different products that it ships to customers throughout the United States. Delivery times are not a driving factor in the decision as to which type of carrier to use (rail, plane, or truck) to deliver the product. However, breakage cost is very expensive, and Frasier would like to select a mode of delivery that reduces the amount of product breakage. To help it reach a decision, the managers have decided to examine the dollar amount of breakage incurred by the three alternative modes of transportation under consideration. Because each product's fragility is different, the executives conducting the study wish to control for differences due to type of product. The company randomly assigns each product to each carrier and monitors the dollar breakage that occurs over the course of 100 shipments. The dollar breakage per shipment (to the nearest dollar) is as follows:Was Frasier and Company correct in its decision to block for type of product? Conduct the appropriate hypothesis test using a level of significance of 0.01.A) Because F = 32.12 > Fα=0.01= 9.78, reject the null hypothesis. Thus, based on these sample data we conclude that blocking is effective.B) Because F = 28.14 > Fα=0.01= 7.63, reject the null hypothesis. Thus, based on these sample data we conclude that blocking is effective.C) Because F = 32.12 > Fα=0.01= 9.78, do not reject the null hypothesis. Thus, based on these sample data we conclude that blocking is not effective.D) Because F = 28.14 > Fα=0.01= 7.63, do not reject the null hypothesis. Thus, based on these sample data we conclude that blocking is not effective.

Q: The following sample data were recently collected in the course of conducting a randomized block analysis of variance. Based on these sample data, what conclusions should be reached about blocking effectiveness and about the means of the three populations involved? Test using a significance level equal to 0.05.A) Because F = 0.4195 < critical F = 4.103, we do not reject the null hypothesis and conclude that the three populations may have the same mean value.B) Because F = 0.4195 < critical F = 4.103, we reject the null hypothesis and conclude that the three populations do not have the same mean value.C) Because F = 0.1515 < critical F = 4.103, we do not reject the null hypothesis and conclude that the three populations may have the same mean value.D) Because F = 0.1515 < critical F = 4.103, we reject the null hypothesis and conclude that the three populations do not have the same mean value.

Q: Consider the following:Test the main hypothesis of interest using α = 0.05A) Because F = 15.65 > critical F = 3.0, we reject the null hypothesis and conclude that the four populations do not have the same mean.B) Because F = 15.65 > critical F = 3.0, we do not reject the null hypothesis and conclude that the four populations have the same mean.C) Because F = 125.82 > critical F = 3.0, we reject the null hypothesis and conclude that the four populations do not have the same mean.D) Because F = 125.82 > critical F = 3.0, we do not reject the null hypothesis and conclude that the four populations have the same mean.

Q: Consider the following:Test to determine whether blocking is effective using an alpha level equal to 0.05A) Because F = 14.81 > critical F = 2.5, we do not reject the null hypothesis and conclude that blocking is not effective.B) Because F = 14.81 > critical F = 2.5, we reject the null hypothesis and conclude that blocking is effective.C) Because F = 112.79 > critical F = 2.5, we do not reject the null hypothesis and conclude that blocking is not effective.D) Because F = 112.79 > critical F = 2.5, we reject the null hypothesis and conclude that blocking is effective.

Q: Consider the following:How many populations are involved in this test?A) 4B) 2C) 5D) 3

Q: Consider the following:How many blocks were used in this study?A) 10B) 9C) 7D) 8

Q: A study was conducted to determine if differences in new textbook prices exist between on-campus bookstores, off-campus bookstores, and Internet bookstores. To control for differences in textbook prices that might exist across disciplines, the study randomly selected 12 textbooks and recorded the price of each of the 12 books at each of the three retailers. You may assume normality and equal-variance assumptions have been met. The partially completed ANOVA table based on the study's findings is shown here:Based on the study's findings, can it be concluded that there is a difference in the average price of textbooks across the three retail outlets? Conduct the appropriate hypothesis test at the alpha = 0.10 level of significance.A) F = 0.0411 < Fα=0.10= 2.56, reject the null hypothesis. Thus, based on these sample data we can conclude that there is a difference in textbook prices at the three different types of retail outlets.B) F = 0.0411 < Fα=0.10= 2.56, do not reject the null hypothesis. Thus, based on these sample data we cannot conclude that there is a difference in textbook prices at the three different types of retail outlets.C) F = 0.031 < Fα=0.10= 2.56, reject the null hypothesis. Thus, based on these sample data we can conclude that there is a difference in textbook prices at the three different types of retail outlets.D) F = 0.031 < Fα=0.10= 2.56, do not reject the null hypothesis. Thus, based on these sample data we cannot conclude that there is a difference in textbook prices at the three different types of retail outlets.

Q: The Lottaburger restaurant chain in central New Mexico is conducting an analysis of its restaurants, which take pride in serving burgers and fries to go faster than the competition. As a part of its analysis, Lottaburger wants to determine if its speed of service is different across its four outlets. Orders at Lottaburger restaurants are tracked electronically, and the chain is able to determine the speed with which every order is filled. The chain decided to randomly sample 20 orders from each of the four restaurants it operates. The speed of service for each randomly sampled order was noted and is contained in the file Lottaburger. At the alpha = 0.05 level of service, can Lottaburger conclude that the speed of service is different across the four restaurants in the chain? A) Since F = 18.418 > Fα=0.05= 2.725, reject the null hypothesis. Based on these sample data we can conclude that the average service time is different across the four restaurants in the chain. B) Since F = 22.666 > Fα=0.05= 2.725, reject the null hypothesis. Based on these sample data we can conclude that the average service time is different across the four restaurants in the chain. C) Since F = 22.666 > Fα=0.05= 2.725, do not reject the null hypothesis. Based on these sample data there is not sufficient evidence to conclude that the average service time is different across the four restaurants in the chain. D) Since F = 18.418 > Fα=0.05= 2.725, do not reject the null hypothesis. Based on these sample data there is not sufficient evidence to conclude that the average service time is different across the four restaurants in the chain.

Q: Damage to homes caused by burst piping can be expensive to repair. By the time the leak is discovered, hundreds of gallons of water may have already flooded the home. Automatic shutoff valves can prevent extensive water damage from plumbing failures. The valves contain sensors that cut off water flow in the event of a leak, thereby preventing flooding. One important characteristic is the time (in milliseconds) required for the sensor to detect the water leak. Sample data obtained for four different shutoff valves are contained in the file entitled Waterflow. Use the Tukey-Kramer multiple comparison technique to discover any differences in the average detection time. Use a significance level of 0.05. A) The confidence intervals indicate that there is not sufficient evidence to conclude that the average detection time for valve 1, 2, and 3 differ. There is, however, enough evidence to indicate that the average detection time for valve 4 is larger than the other three means. B) The confidence intervals indicate that there is not sufficient evidence to conclude that the average detection time for valve 1, 2, and 4 differ. There is, however, enough evidence to indicate that the average detection time for valve 3 is larger than the other three means. C) The confidence intervals indicate that there is not sufficient evidence to conclude that the average detection time for valve 2 and 4 differ. There is, however, enough evidence to indicate that the average detection time for valve 1 and 3 are larger than the other two means. D) All mean detection times are equal.

Q: In conjunction with the housing foreclosure crisis of 2009, many economists expressed increasing concern about the level of credit card debt and efforts of banks to raise interest rates on these cards. The banks claimed the increases were justified. A Senate subcommittee decided to determine if the average credit card balance depends on the type of credit card used. Under consideration are Visa, MasterCard, Discover, and American Express. The sample sizes to be used for each level are 25, 25, 26, and 23, respectively. State the number of degrees of freedom available for determining the total variation. A) 93 B) 95 C) 98 D) 97

Q: Given the following sample dataBased on the computations for the within- and between-sample variation, develop the ANOVA table and test the appropriate null hypothesis using alpha= 0.05. Use the p-value approach.A) Since p-value = 0.0678 > 0.05 reject H0and conclude that at least two population means are different.B) Since p-value = 0.000136 < 0.05 reject H0and conclude that at least two population means are different.C) Since p-value = 0.0678 > 0.05 accept H0and conclude that all population means are the same.D) Since p-value = 0.000136 < 0.05 accept H0and conclude that all population means are the same.

Q: Respond to the following questions using this partially completed one-way ANOVA table:Based on the analysis of variance F-test, what conclusion should be reached regarding the null hypothesis? Test using alpha = 0.05.A) Since 11.1309 > 2.9467 accept H0and conclude that all population means are the same.B) Since 2.9467 > 11.1309 accept H0and conclude that all population means are the same.C) Since 11.1309 > 2.9467 reject H0and conclude that at least two populations means are different.D) Since 2.9467 > 11.1309 reject H0and conclude that at least two populations means are different.

Q: Respond to the following questions using this partially completed one-way ANOVA table:Fill in the ANOVA table with the missing values.A) SSB = 483, MSB = 161, F-ratio = 11.1309, Within Samples df = 28, MSW = 14.464B) SSB = 483, MSB = 161, F- ratio = 8.1629, Within Samples df = 28, MSW = 14.464C) SSB = 483, MSB = 161, F-ratio = 8.1629, Within Samples df = 25, MSW = 14.464D) SSB = 504, MSB = 161, F-ratio = 8.1629, Within Samples df = 28, MSW = 14.464

Q: Respond to the following questions using this partially completed one-way ANOVA table:How many different populations are being considered in this analysis?A) 2B) 4C) 6D) 5

Q: Respond to the following questions using this partially completed one-way ANOVA table:Source of Variation SS df MS F-ratioHow many different populations are being considered in this analysis?A) 5B) 8C) 7D) 6

Q: A manager is interested in testing whether three populations of interest have equal population means. Simple random samples of size 10 were selected from each population. The following ANOVA table and related statistics were computed:Conduct the appropriate test of the null hypothesis assuming that the populations have equal variances and the populations are normally distributed. Use a 0.05 level of significance.A) Using the F test approach, because F = 3.354 < critical F = 9.84, we reject the null hypothesis and conclude that the population means are not all equal.B) Using the F test approach, because F = 3.354 < critical F = 9.84, we do not reject the null hypothesis and conclude that the population means are all equal.C) Using the F test approach, because F = 9.84 > critical F = 3.35, we reject the null hypothesis and conclude that the population means are not all equal.D) Using the F test approach, because F = 9.84 > critical F = 3.35, we do not reject the null hypothesis and conclude that the population means are all equal.

Q: A start-up cell phone applications company is interested in determining whether house-hold incomes are different for subscribers to three different service providers. A random sample of 25 subscribers to each of the three service providers was taken, and the annual household income for each subscriber was recorded. The partially completed ANOVA table for the analysis is shown here:Based on the sample results, can the start-up firm conclude that there is a difference in household incomes for subscribers to the three service providers? You may assume normal distributions and equal variances. Conduct your test at the alpha= 0.10 level of significance. Be sure to state a critical F-statistic, a decision rule, and a conclusion.A) H0 : 1= 2= 3HA: Not all populations have the same meanF = MSB/MSW = 1,474,542,579/87,813,791 = 16.79Because the F test statistic = 16.79 > Fα= 2.3778, we do reject the null hypothesis based on these sample data.B) H0 : 1= 2= 3HA : Not all populations have the same meanF = MSB/MSW = 87,813,791 /1,474,542,579= 0.060Because the F test statistic = 0.060 < Fα= 2.3778, we do not reject the null hypothesis based on these sample data.C) H0 : 1= 2= 3HA : Not all populations have the same meanF = SSW/MSW = 6,322,592,933/87,813,791 = 72Because the F test statistic = 72 > Fα= 2.3778, we do reject the null hypothesis based on these sample data.D) H0 : 1= 2= 3HA : Not all populations have the same meanF = SSW/MSW = 6,322,592,933/1,474,542,579= 4.28Because the F test statistic = 4.28 > Fα= 2.3778, we do reject the null hypothesis based on these sample data

Q: A two-factor analysis of variance is conducted to test the effect that price and advertising have on sales of a particular brand of bottled water. Each week a combination of particular levels of price and advertising are used and the sales amount is recorded. The computer results are shown below.Based on the results above and a 0.05 level of significance, which of the following is correct?A) There is no interaction between price and advertising, so results for individual factors may be misleading.B) There is interaction between price and advertising, so the above results for individual factors may be misleading.C) There is no interaction between price and advertising, and both factors significantly affect sales.D) There is interaction between price and advertising, so the above results conclusively show that both factors affect price.

Q: A two-factor analysis of variance is conducted to test the effect the price and advertising have on sales of a particular brand of bottled water. Each week a combination of particular levels of price and advertising are used and the sales level is recorded. The computer results are shown below.Based on the results above, which of the following is correct?A) 1 level of advertising and 2 levels of price were used.B) 3 levels of adverting and 2 levels of price were used.C) 2 levels of advertising and 3 levels of price were used.D) There were a total of 6 different treatments.

Q: A national car rental company recently conducted a study recently in which cars with automatic and standard transmissions (factor A-Sample) were rented to male and female customers (factor B-Columns). Three customers in each category were randomly selected and the miles driven per day was recorded as follows:Based on these sample data, and alpha = .05, which of the following statements is true?A) The means for factor A are significantly different.B) There is no significant interaction between factors A and B.C) The means for factor B are significantly different.D) All of the above statements are true.

Q: Considering the following printout for a two-factor ANOVA study, which of the following is the number of replications used?A) 2B) 5C) 4D) Can't be determined without more information.

Q: Considering the following printout for a two-factor ANOVA design, which of the following is a proper conclusion to reach?A) There is no significant interaction between the two factors.B) The levels of factor A (Sample) have significantly different means.C) The levels of factor B (Columns) have significantly different means.D) The total number of observations is 47.

Q: Considering the following printout from a two-factor ANOVA design, how many levels of factor A (Sample) were there in this study?A) 4B) 3C) 2D) 6

Q: Which of the following is the minimum number of required replications per cell for a two-factor ANOVA design if you plan to test for interactive effects between factors A and B? A) 3 B) 1 C) 2 D) 5

Q: Which type of ANOVA can include interaction? A) One-way B) Randomized complete block C) Two-factor D) All types of ANOVA

Q: Which of the following is NOT one of the assumptions required by the randomized block design? A) The populations are normally distributed. B) The populations have equal means. C) The observations within samples are independent. D) The data measurement must be interval or ratio level.

Q: A car company is interested in testing to see whether the mean miles that a car engine will last without changing oil is the same or different depending on which brand of oil is used. The engineers also wish to control for the type of transmission (manual or automatic) that is used. To conduct this test, the car company obtains enough engines so that all four oil brands can be tested in a design that involves no replication. The following data reflect the miles the engine lasted until problems were encountered. Data are in thousands of miles.Assuming that the hypothesis tests are conducted using a significance level equal to 0.05, which of the following statements is true?A) Based on the data, Oil 1 and Oil 3 give statistically more miles on average than do the other two oils.B) The type of transmission does seem to have an impact on the mean miles that an engine will last.C) The F-critical value for testing whether blocking is effective is 10.128.D) All of the above are true.

Q: A test is conducted to compare three different income tax software packages to determine whether there is any difference in the average time it takes to prepare income tax returns using the three different software packages. Ten different persons' income tax returns are done by each of the three software packages and the time is recorded for each. The computer results are shown below.Assuming that the hypothesis tests are conducted using a significance level equal to 0.05, the Fisher's LSD value for multiple comparisons is:A) approximately 0.4985.B) about 0.91.C) approximately 1.91.D) about 0.5387.

Q: A large orchard owner in the state of Washington is interested in determining whether the mean number of bushels of peaches per acre is the same or different depending on the type of tree that is used. He also thinks that production may be affected by the type of fertilizer that is used. To test, he has set up a test in which a one-acre plot of peach trees with a combination each of 5 varieties and 3 fertilizer types are studied. The following data reflect the number of bushels of peaches on each acre plot.Assuming that the hypothesis tests will be conducted using an alpha equal 0.05 level, what is the value of the Fisher's LSD critical value for doing the multiple comparisons?A) Approximately 16.78B) About 11.30C) Approximately 186.7D) Need to know the number of trees planted on each acre.

Q: A golf ball manufacturer has three dimple patterns it is interested in analyzing to see whether one results in longer driving distances. However, it also wishes to control for the material the ball is made from since it believes that the material might affect driving distance. Four materials can be used. The following data represent the results of tests in which each combination of dimple pattern and cover material were used and the length of the ball hit by a robot has been recorded. The test will be conducted using an alpha = 0.05 level.Given these data, what is the value of Fisher's Least Significant Difference critical value?A) Approximately 19.06B) 2.4469C) About 7.65D) None of the above

Q: A major consumer group recently undertook a study to determine whether automobile customers would rate the quality of cars differently whether they were manufactured in the U.S., Europe, or Japan. To conduct this test, a sample of 20 individuals was asked to look at mid-range model cars made in each of the three countries. The individuals in the sample were then asked to provide a rating for each car on a scale of 1 to 1000. The following computer output resulted, and the tests were conducted using a significance level equal to 0.05.ANOVA: Two-Factor Without ReplicationBased upon the data, which of the following statements is true?A) Blocking was effective.B) Blocking was not effective.C) The primary null hypothesis should not be rejected.D) The averages for the 20 people are not all the same.

Q: Recently, a department store chain was interested in determining if there was a difference in the mean number of customers who enter the three stores in Seattle. The analysts set up a study in which the number of people entering the stores was counted depending on whether the day of the week was Saturday, Sunday, or a weekday. The following data were collected:Given this format and testing using an alpha level equal to 0.05, which of the following statements is true?A) The total degrees of freedom is 9.B) The between blocks degrees of freedom equals 8.C) The between samples degrees of freedom equals 3.D) The within sample degrees of freedom equals 4.

Q: In a randomized complete block design analysis of variance, which of the following correctly describes the number of degrees of freedom associated with the between sum of squares? A) One less than the number of populations involved B) One less than the number of blocks C) One less than the combined sample size in the experiment D) One less than the total number of observations

Q: In a randomized complete block design analysis of variance, how many factors are there to be analyzed? A) Depends on the sample size in each treatment B) One factor, but multiple levels C) Two factors D) Can't be determined without additional information

Q: Which of the following describes a treatment in a randomized complete block analysis of variance? A) A treatment is a combination of one level of each factor. B) A treatment is a level associated with each factor of the experiment. C) A treatment is another term for the data that are collected in the experiment. D) A treatment is considered to be the analysis that is performed on the sample data.

Q: To test the mileage efficiency of three new car models, random samples of various sample sizes were selected from each of the three cars and the mpg data obtained are shown below.Based on the sample date, one can conclude thatA) all three car models have the same mean mpg.B) at least two car models have different mpgs.C) Model C has a higher mpg than Model A.D) None of the above