# MATHEMATICS

**1.** TABLE 11-3

A realtor wants to compare the average sales-to-appraisal ratios of residential properties sold in four neighborhoods (A, B, C, and D). Four properties are randomly selected from each neighborhood and the ratios recorded for each, as shown below.

A: 1.2, 1.1, 0.9, 0.4 C: 1.0, 1.5, 1.1, 1.3

B: 2.5, 2.1, 1.9, 1.6 D: 0.8, 1.3, 1.1, 0.7

Interpret the results of the analysis summarized in the following table:

MA

Referring to Table 11-3, the among-group degrees of freedom is

[removed]A) 3.

[removed]B) 4.

[removed]C) 16.

[removed]D) 12.

**2.** TABLE 11-3

A realtor wants to compare the average sales-to-appraisal ratios of residential properties sold in four neighborhoods (A, B, C, and D). Four properties are randomly selected from each neighborhood and the ratios recorded for each, as shown below.

A: 1.2, 1.1, 0.9, 0.4 C: 1.0, 1.5, 1.1, 1.3

B: 2.5, 2.1, 1.9, 1.6 D: 0.8, 1.3, 1.1, 0.7

Interpret the results of the analysis summarized in the following table:

Referring to Table 11-3, the within-group sum of squares is

[removed]A) 1.0606.

[removed]B) 4.3644.

[removed]C) 1.1825.

[removed]D) 3.1819.

**3.** A completely randomized design

[removed]A) has one factor and one block and multiple values.

[removed]B) can have more than one factor, each with several treatment groups.

[removed]C) has one factor and one block.

[removed]D) has only one factor with several treatment groups.

**4.** The *F* test statistic in a one-way ANOVA is

[removed]A) MSA/MSW.

[removed]B) SSA/SSW.

[removed]C) MSW/MSA.

[removed]D) SSW/SSA.

**5.** TABLE 11-2

An airline wants to select a computer software package for its reservation system. Four software packages (1, 2, 3, and 4) are commercially available. The airline will choose the package that bumps as few passengers, on the average, as possible during a month. An experiment is set up in which each package is used to make reservations for 5 randomly selected weeks. (A total of 20 weeks was included in the experiment.) The number of passengers bumped each week is obtained, which gives rise to the following Excel output:

ANOVA

Referring to Table 11-2, the within groups degrees of freedom is

[removed]A) 19.

[removed]B) 16.

[removed]C) 4.

[removed]D) 3.

**6.** TABLE 11-5

A physician and president of a Tampa Health Maintenance Organization (HMO) are attempting to show the benefits of managed health care to an insurance company. The physician believes that certain types of doctors are more cost-effective than others. One theory is that Primary Specialty is an important factor in measuring the cost-effectiveness of physicians. To investigate this, the president obtained independent random samples of 20 HMO physicians from each of 4 primary specialties – General Practice (GP), Internal Medicine (IM), Pediatrics (PED), and Family Physicians (FP) – and recorded the total charges per member per month for each. A second factor which the president believes influences total charges per member per month is whether the doctor is a foreign or USA medical school graduate. The president theorizes that foreign graduates will have higher mean charges than USA graduates. To investigate this, the president also collected data on 20 foreign medical school graduates in each of the 4 primary specialty types described above. So information on charges for 40 doctors (20 foreign and 20 USA medical school graduates) was obtained for each of the 4 specialties. The results for the ANOVA are summarized in the following table.

Referring to Table 11-5, what assumption(s) need(s) to be made in order to conduct the test for differences between the mean charges of foreign and USA medical school graduates?

[removed]A) The charges in each group of doctors sampled are drawn from normally distributed populations.

[removed]B) The charges in each group of doctors sampled are drawn from populations with equal variances.

[removed]C) There is no significant interaction effect between the area of primary specialty and the medical school on the doctors’ mean charges.

[removed]D) All of the above are necessary assumptions.

**7.** TABLE 12-2

Many companies use well-known celebrities as spokespersons 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.

Referring to Table 12-2, at 5% level of significance, the critical value of the test statistic is:

[removed]A) 3.8415

[removed]B) 13.2767

[removed]C) 5.9914

[removed]D) 9.4877

**8.** TABLE 12-18

An agronomist wants to compare the crop yield of 3 varieties of chickpea seeds. She plants all 3 varieties of the seeds on each of 5 different patches of fields. She then measures the crop yield in bushels per acre. Treating this as a randomized block design, the results are presented in the table that follows.

Referring to Table 12-18, the null hypothesis for the Friedman rank test is

[removed]A) *H*_{0}: *M*_{Field1} = *M*_{Field2} = *M*_{Field3} = *M*_{Field4} = *M*_{Field5}

[removed]B) *H*_{0}: *M*_{Smith} = *M*_{Walsh} = *M*_{Trevor}

[removed]C) *H*_{0}: *μ*_{Field1} = *μ*_{Field2} = *μ*_{Field3} = *μ*_{Field4} = *μ*_{Field5}

[removed]D) *H*_{0}: *μ*_{Smith} = *μ*_{Walsh} = *μ*_{Trevor}

**9.** TABLE 12-10

Parents complain that children read too few storybooks and watch too much television nowadays. A survey of 1,000 children reveals the following information on average time spent watching TV and average time spent reading storybooks.

Average time spent reading story books

Average time spent watching TV | Less than 1 hour |
Between 1 and 2 hours |
More than 2 hours |

Less than 2 hours | 90 | 85 | 130 |

More than 2 hours | 655 | 32 | 8 |

Referring to Table 12-10, we want to test whether there is any relationship between average time spent watching TV and average time spent reading storybooks. Suppose the value of the test statistic was 164 (which is not the correct answer) and the critical value was 19.00 (which is not the correct answer), then we could conclude that

[removed]A) there is no connection between time spent reading storybooks and time spent watching TV.

[removed]B) there is connection between time spent reading storybooks and time spent watching TV.

[removed]C) more time spent watching TV leads to less time spent reading storybooks.

[removed]D) more time spent reading storybooks leads to less time spent watching TV.

**10.** TABLE 12-3

A computer used by a 24-hour banking service is supposed to randomly assign each transaction to one of 5 memory locations. A check at the end of a day’s transactions gave the counts shown in the table for each of the 5 memory locations, along with the number of reported errors.

The bank manager wanted to test whether the proportion of errors in transactions assigned to each of the 5 memory locations differ.

Referring to Table 12-3, the critical value of the test statistic at 1% level of significance is:

[removed]A) 13.2767

[removed]B) 7.7794

[removed]C) 23.2093

[removed]D) 20.0902