Q&A Hero

Q: A study was recently done in the United States in which car owners were asked to indicate whether their most recent car purchase was a U.S. car, a German car, or a Japanese car. The people in the survey were divided by geographic region in the United States. The following data were recorded. US Japanese German East Coast 200 200 50 Central 250 100 20 West Coast 80 300 40 Given this situation, to test whether the car origin is independent of the geographical location of the buyer, the critical value for alpha = .10 is 14.6837.

Q: A study was recently done in the United States in which car owners were asked to indicate whether their most recent car purchase was a U.S. car, a German car, or a Japanese car. The people in the survey were divided by geographic region in the United States. The following data were recorded. US Japanese German East Coast 200 200 50 Central 250 100 20 West Coast 80 300 40 Given this situation, to test whether the car origin is independent of the geographical location of the buyer, the sum of the expected cell frequencies will equal 1,240.

Q: A study was recently done in the United States in which car owners were asked to indicate whether their most recent car purchase was a U.S. car, a German car, or a Japanese car. The people in the survey were divided by geographic region in the United States. The following data were recorded. US Japanese German East Coast 200 200 50 Central 250 100 20 West Coast 80 300 40 Given this situation, the null hypothesis to be tested is that the car origin is dependent on the geographical location of the buyer.

Q: A study was recently done in the United States in which car owners were asked to indicate whether their most recent car purchase was a U.S. car, a German car, or a Japanese car. The people in the survey were divided by geographic region in the United States. The following data were recorded. US Japanese German East Coast 200 200 50 Central 250 100 20 West Coast 80 300 40 Given this situation, the sample size used in this study was nine.

Q: In order to apply the chi-square contingency methodology for quantitative variables, we must first break the quantitative variable down into discrete categories.

Q: A cell phone company wants to determine if the use of text messaging is independent of age. The following data has been collected from a random sample of customers. Regularly use text messaging Do not regularly use text messaging Under 21 82 38 21-39 57 34 40 and over 6 8 Using this data, if we wish to test whether the preferred news source is independent of age using a 0.05 level of significance, the critical value is 5.9915.

Q: A study was recently conducted in which people were asked to indicate which news medium was their preferred choice for national news. The following data were observed: radio television newspaper under 21 30 50 5 21-40 20 25 30 41 and over 30 30 50 Given this data, if we wish to test whether the preferred news source is independent of age, for an alpha = .05 level, the test statistic is computed to be approximately 40.70.

Q: A study was recently conducted in which people were asked to indicate which news medium was their preferred choice for national news. The following data were observed: radio television newspaper under 21 30 50 5 21-40 20 25 30 41 and over 30 30 50 Given this data, if we wish to test whether the preferred news source is independent of age, for an alpha = .05 level, the critical value from the chi-square table is 9.4877.

Q: A study was recently conducted in which people were asked to indicate which news medium was their preferred choice for national news. The following data were observed: radio television newspaper under 21 30 50 5 21-40 20 25 30 41 and over 30 30 50 Given this data, if we wish to test whether the preferred news source is independent of age, for an alpha = .05 level, the critical value from the chi-square table is based on 8 degrees of freedom.

Q: A study was recently conducted in which people were asked to indicate which news medium was their preferred choice for national news. The following data were observed: radio television newspaper under 21 30 50 5 21-40 20 25 30 41 and over 30 30 50 Given this data, if we wish to test whether the preferred news source is independent of age, the cell with the largest expected cell frequency is also the cell with the largest observed frequency.

Q: A study was recently conducted in which people were asked to indicate which news medium was their preferred choice for national news. The following data were observed: radio television newspaper under 21 30 50 5 21-40 20 25 30 41 and over 30 30 50 Given this data, if we wish to test whether the preferred news source is independent of age with an alpha equal to .05, the critical value will be a chi-square value with 9 degrees of freedom.

Q: A cell phone company wants to determine if the use of text messaging is independent of age. The follow data has been collected from a random sample of their customers. Regularly use text messaging Do not regularly use text messaging Under 21 82 38 21-39 57 34 40 and over 6 83 Using the data above, in order to test for the independence of age and the use of text messaging, the expected value for the "under 21 and regularly use text messaging" cell is 82.

Q: A study was recently conducted in which people were asked to indicate which new medium was their preferred choice for national news. The following data were observed: radio television newspaper under 21 30 50 5 21-40 20 25 30 41 and over 30 30 50 Given this data, if we wish to test whether the preferred news source is independent of age, the expected frequency in the cell, radiounder 21 cell is 30.

Q: In conducting a test of independence for a contingency table that has 4 rows and 3 columns, the number of degrees of freedom is 11.

Q: When the variables of interest are both categorical and the decision maker is interested in determining whether a relationship exists between the two, a statistical technique known as contingency analysis is useful.

Q: A survey was recently conducted in which males and females were asked whether they owned a laptop personal computer. The following data were observed: Males Females Have Laptop 120 70 No Laptop 50 60 Given this information, if an alpha level of .05 is used, the test statistic for determining whether having a laptop is independent of gender is approximately 14.23.

Q: A survey was recently conducted in which males and females were asked whether they owned a laptop personal computer. The following data were observed: Males Females Have Laptop 120 70 No Laptop 50 60 Given this information, if an alpha level of .05 is used, the sum of the expected cell frequencies will be equal to the sum of the observed cell frequencies.

Q: A survey was recently conducted in which males and females were asked whether they owned a laptop personal computer. The following data were observed: Males Females Have Laptop 120 70 No Laptop 50 60 Given this information, if an alpha level of .05 is used, the critical value for testing whether the two variables are independent is x2 = 3.8415.

Q: A survey was recently conducted in which males and females were asked whether they owned a laptop personal computer. The following data were observed: Males Females Have Laptop 120 70 No Laptop 50 60 Given this information, if having a laptop is independent of gender, the expected number of males with laptops in this survey is 150.

Q: A survey was recently conducted in which males and females were asked whether they owned a laptop personal computer. The following data were observed: Males Females Have Laptop 120 70 No Laptop 50 60 Given this information, the sample size in the survey was 300 people.

Q: Managers use contingency analysis to determine whether two categorical variables are independent of each other.

Q: Contingency analysis is used only for numerical data.

Q: Contingency analysis helps to make decisions when multiple proportions are involved.

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 H0 should 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: 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 sum of the expected frequencies over the six days cannot be determined without seeing the actual sample data.

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 expected number that will arrive on Monday is about 33.33.

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: By combining cells we guard against having an inflated test statistic that could have led us to incorrectly accept the null hypothesis.

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: 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, and without regard to whether there is a need to combine cells due to expected cell frequencies, the critical value for testing whether the distribution is Poisson with a mean of 3.5 per hour at an alpha level of .05 is x2 = 15.5073.

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. X Frequency 0 28 1 38 2 22 3 7 4 or 5 5 Total 100 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.

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. X Frequency 0 28 1 38 2 22 3 7 4 or 5 5 Total 100 Given this information the expected number of days on which exactly 1 machine breaks down is 40.96.

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. X Frequency 0 28 1 38 2 22 3 7 4 or 5 5 Total 100 Given this information, assuming that all expected values are sufficiently large to use the classes as shown above, the critical value for testing the hypothesis will be based on 5 degrees of freedom.

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: 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. The degrees of freedom for the test will be 7.

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: Recently a survey was conducted involving customers of a fitness center in Dallas, Texas. Participants were asked to indicate how often they use the club by checking one of the following categories: 0-1 time per week; 2-3 times per week; 4-5 times per week; more than 5 times. The following data show how males and females responded to this question.0-12-34-5over 5Males41615020Females109896045One of the purposes of the survey was to determine whether there is a relationship between the gender of the customer and the number of visits made each week.a. State the appropriate null and alternative hypothesis.b. What test procedure is appropriate to use to conduct this test?c. Conduct the hypothesis test using an alpha = .05 level.

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

Q: The Bradfield Container Company makes "cardboard" boxes for commercial use (i.e., pizza boxes). One of the big issues for the company is the set-up time required to change over from one order to the next. At one particular machine, the set-up time is thought to be uniformly distributed between 10 and 21 minutes. To test whether this is true or not, a random sample of 180 set-ups on this machine was selected with set-up time rounded to the nearest two-minute intervals. The following results occurred:Set-up Time Frequency10-11 minutes 1312-13 minutes 2314-15 minutes 4016-17 minutes 4418-19 minutes 4020-21 minutes 20a. What are the appropriate null and alternative hypothesis to be tested?b. Based on the null and alternative hypotheses stated in part a, determine the expected frequencies for each set-up time category.c. Assuming that we wish to conduct the hypothesis test at the .05 level, what is the critical value that should be used?d. Compute the test statistic and carry out the hypothesis test.

Q: Explain the basic logic behind the chi-square goodness-of-fit test.

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 independent C) 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 as At 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: 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, which of the following values is the correct value of the test statistic? A) Approximately 9.48 B) Nearly 23.0 C) About 3.84 D) Approximately 5.94

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 hypothesis test is to be conducted using a .05 level of significance, which of the following is correct critical value? A) 9.4877 B) 3.8415 C) 1.96 D) 7.8147

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, which test would be used to properly analyze the data in this experiment? A) x2 test for independence in a two-way contingency table B) x2 test for equal proportions in a one-way table C) ANOVA F-test for main treatment effect D) x2 goodness-of-fit test

Q: A personal computer assembly company is interested in studying the time it takes to assemble a computer under different conditions. Specifically, two factors are to be controlled: Factor A (Sample) is the number of individual work stations in the line and Factor B (Columns) is whether the assembly was done on day shift, swing shift, or graveyard shift. The following data were obtained in the study where the response variable is the time required to assemble the PC from start to final test. Carry out the appropriate statistical tests using an alpha equal .05 level.

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

Q: Suppose that you have just performed an experiment using the randomized complete block analysis of variance design and the output from Excel is shown as follows. The primary null hypothesis to be tested involves whether the means of the three groups are equal. ANOVA: Two-Factor Without Replication Summary Count Sum Average Variance Block 1 3 450 150 468.9 Block 2 3 210 70 72.9 Block 3 3 610 203.3333 633.3333 Block 4 3 700 233.3333 3333.333 Group 1 4 560 140.0 5066.667 Group 2 4 810 202.5 8425 Group 3 4 600 150 3266.667 ANOVA Source of Variation SS df MS F p-value F-crit Rows 46158.33 3 15386.11 22.4251 0.001163 4.757055 Columns 9016.667 2 4508.333 6.57085 0.030797 5.143249 Error 4116.667 6 686.1111 Total 5929.67 11 Using a significance level of 0.05, what conclusions should be made about the three groups? Which groups have different means? Discuss.

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: A real estate broker is interested in determining whether there is a difference in the mean number of days a home stays on the market before selling based on which area of the city it is located in. However, she is also concerned that the price of the house may be an issue in determining how long it takes to sell a house, so she wants to control for this. To carry out the test, she plans to randomly select one house from each part of the city in each price range. The following data show the number of days for the sample of houses selected. East West North South under $70,000 42 60 29 50 $70,000 < $90,000 40 70 40 37 $90,000 < $120,000 50 80 60 30 $120,000 < $180,000 30 40 56 40 $180,000 and over 56 33 40 20 Using a significance level equal to 0.05, determine whether the broker was justified in controlling for house prices. Be sure to indicate what type of statistical test should be used and why.

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

Q: A study has been conducted to determine whether the mean spending for recreational activities during the month of August differs for residents of three cities. Random samples of 30 people were selected from each city and their spending on recreation was recorded during August. The following output was generated using Excel: ANOVA: Single Factor SUMMARY Groups Count Sum Average Variance City 1 30 7897.179 236.2393 3334.11 City 2 30 10322.1 344.0701 2201.818 City 3 30 6045.102 201.5034 2215.919 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 306701.8 2 153350.9 59.3475 5.54E-17 3.101292 Within Groups 224803.5 87 2583.949 Total 531505.4 89 The Excel output shows that the null hypothesis of equal means should be rejected. Given this, perform the appropriate method for determining which population means are different. Conduct the test using an alpha = 0.05 level.

Q: A study has been conducted to determine whether the mean spending for recreational activities during the month of August differs for residents of three cities. Random samples of 30 people were selected from each city and their spending on recreation was recorded during August. The following output was generated using Excel: ANOVA: Single Factor SUMMARY Groups Count Sum Average Variance City 1 30 7897.179 236.2393 3334.11 City 2 30 10322.1 344.0701 2201.818 City 3 30 6045.102 201.5034 2215.919 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 306701.8 2 153350.9 59.3475 5.54E-17 3.101292 Within Groups 224803.5 87 2583.949 Total 531505.4 89 Based on the information provided, should we conclude that the three populations (cities) have equal mean spending during August? Test at the 0.05 level of significance.

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: Source SS df MS F-Ratio Factor A 162.79 4 Factor B 28.12 AB Interaction 262.31 12 Error ________ ___ Total 1,298.74 84 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: Source SS df MS F-Ratio Factor A 162.79 4 Factor B 28.12 AB Interaction 262.31 12 Error ________ ___ Total 1,298.74 84 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: Source SS df MS F-Ratio Factor A 162.79 4 Factor B 28.12 AB Interaction 262.31 12 Error ________ ___ Total 1,298.74 84 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 B B) Reject H0. Conclude that there is sufficient evidence to indicate interaction exists between Factor A and Factor B C) Fail to reject H0. Conclude that there is sufficient evidence to indicate interaction exists between Factor A and Factor B D) 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: Source SS df MS F-Ratio Factor A 162.79 4 Factor B 28.12 AB Interaction 262.31 12 Error ________ ___ Total 1,298.74 84 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 = 12.143 B) 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 = 12.143 C) 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 = 12.143 D) 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: 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 these sample data, can you conclude the three grocery stores have different sample means? Test using a significance level of 0.05. Use a p-value approach. A) Since the p-value of 2.68E - 10 < 0.05 reject H0 and conclude that at least two means are different. B) Since the p-value of 2.68E-10 < 0.05 do not reject H0 and conclude that means are all the same. C) Since the p-value of 0.027 < 0.05 reject H0 and conclude that at least two means are different. D) Since the p-value of 0.027 < 0.05 do not reject H0 and conclude that means are all the same.

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 a significance level of 0.05 and these sample data, test to determine whether blocking was necessary in this example. Use a test-statistic approach. A) Since 952.6155 > 1.8673 do not reject H0 and conclude that there is an indication that blocking was not effective. B) Since 952.6155 > 1.8673 reject H0 and conclude that there is an indication that blocking was effective. C) Since 102.2912 > 1.8673 do not reject H0 and conclude that there is an indication that blocking was not effective. D) Since 102.2912 > 1.8673 reject H0 and conclude that there is an indication that blocking was effective.

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.