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KP

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The price of a 40-ounce bottle of rum at a U.S. border duty-free store was US$19.95. At the same time, the price of a 750-mL bottle of similar quality rum in an Ontario liquor store was C$22.95. Using the Ontario price as the base, what is the percent saving (on the unit price) by buying rum at the duty-free store? (1 litre = 35.2 fluid ounces)

On Jul 09, 2024


40.68%
KP

Answered

What does the principle of parsimony suggest?

A) A simple model with fewer independent variables may not produce an effective result.
B) The fewest number of explanatory variables that explain the independent variable need to be included in the model.
C) Good regression models are often based on sound technical analysis.
D) To avoid the problem of multicollinearity,the number of independent variables need to be sufficiently high.

On Jul 05, 2024


B
KP

Answered

In a regression analysis involving 21 observations and 4 independent variables, the following information was obtained.
R2 = .80
s = 5.0
Based on the above information, fill in all the blanks in the following ANOVA table.
Hint: R2 = In a regression analysis involving 21 observations and 4 independent variables, the following information was obtained. R<sup>2</sup> = .80 s = 5.0 Based on the above information, fill in all the blanks in the following ANOVA table. Hint: R<sup>2</sup> =   , but also R<sup>2</sup> = 1 -   .  , but also R2 = 1 - In a regression analysis involving 21 observations and 4 independent variables, the following information was obtained. R<sup>2</sup> = .80 s = 5.0 Based on the above information, fill in all the blanks in the following ANOVA table. Hint: R<sup>2</sup> =   , but also R<sup>2</sup> = 1 -   .  . In a regression analysis involving 21 observations and 4 independent variables, the following information was obtained. R<sup>2</sup> = .80 s = 5.0 Based on the above information, fill in all the blanks in the following ANOVA table. Hint: R<sup>2</sup> =   , but also R<sup>2</sup> = 1 -   .

On Jun 07, 2024


KP

Answered

Multiply the fractions (49) (−52) \left( \frac { 4 } { 9 } \right) \left( - \frac { 5 } { 2 } \right) (94) (25) and write the result in simplest form.

A) −3718- \frac { 37 } { 18 }1837
B) 17\frac { 1 } { 7 }71
C) - 17\frac { 1 } { 7 }71
D) −109- \frac { 10 } { 9 }910
E) 109\frac { 10 } { 9 }910

On Jun 03, 2024


D
KP

Answered

Calculate the periodic payment for the following ordinary annuity.
Calculate the periodic payment for the following ordinary annuity.

On May 08, 2024


$3,324.24
KP

Answered

The research hypothesis posits that there will be a relationship between the number of hours a student studies and their result on a test (H1: rxy ≠ 0).What would be concluded for r(45)= 0.213,p > 0.05?

On May 04, 2024


The obtained value (0.213)is less than the critical value (0.2875),so it cannot be concluded that the relationship between the two variables occurred by something other than chance.Furthermore,the research hypothesis,which posits a relationship between the variables,is also not supported.
KP

Answered

Hill County is divided into three communities whose assessed valuations, determined individually, are shown in the table below, along with the percentage of assessed value to true value. In the answer column on the right, show what each community's assessment should be for the fair sharing of the county's overhead expenses. Round answers to the nearest dollar.
Hill County is divided into three communities whose assessed valuations, determined individually, are shown in the table below, along with the percentage of assessed value to true value. In the answer column on the right, show what each community's assessment should be for the fair sharing of the county's overhead expenses. Round answers to the nearest dollar. ​

On May 03, 2024


KP

Answered

The following ANOVA table shows the results of an experiment used to test how race car drivers and different tracks affect time.The experiment used 3 different drivers on 4 different tracks.Assuming no effect from the interaction between driver and track,test the claim that the three drivers have the same mean time.  Sum ofMean Source  DF  Squares  Square  F-ratio  P-value  Driver 2210.330.729 Tradk 398.2532.7510.92<0.0001 Error 6183 Total 11118.25\begin{array}{l}\begin{array} { l r c c c c } &&\text { Sum of}&{ Mean }\\\text { Source } & \text { DF } & \text { Squares } & \text { Square } & \text { F-ratio } & \text { P-value } \\\text { Driver } & 2 & 2 & 1 & 0.33 & 0.729 \\\text { Tradk } & 3 & 98.25 & 32.75 & 10.92 & < 0.0001 \\\text { Error } & 6 & 18 & 3 & & \\\text { Total } & 11 & 118.25 & & &\end{array}\end{array} Source  Driver  Tradk  Error  Total  DF 23611 Sum of Squares 298.2518118.25Mean Square 132.753 F-ratio 0.3310.92 P-value 0.729<0.0001

On May 01, 2024


H0: γ1 = γ2 = γ3,where γ represents the effect of the driver vs.HA: Not all of the drivers have the same effect.The null hypothesis states that the driver has no effect on the time.The alternative states that it does.The P-value for the driver is 0.729,providing insufficient evidence that the driver has an effect on mean times.
KP

Answered

A hunting supply store is interested in knowing whether three different brands of goose decoys are equally effective in attracting geese.Each goose decoy is randomly assigned to five hunters,and the order of each experiment is also randomized.An experiment consists of a hunter using one goose decoy and waiting fifteen minutes.The number of geese attracted by the decoy is counted.

A) H0: μ1 ≠ μ2 ≠ μ3 vs.the alternative that all decoy brand mean number of geese are equal
B) H0: μ1 ≠ μ2 ≠ μ3 ≠ μ4 ≠ μ5 vs.the alternative that all hunter mean number of geese are equal
C) H0: μ1 = μ2 = μ3 vs.the alternative that not all decoy brand mean number of geese are equal
D) H0: μ1 = μ2 = μ3 = μ4 = μ5 vs.the alternative that not all hunter mean number of geese are equal
E) H0: μbrand ≠ μhunter vs.the alternative that decoy brand and hunter mean number of geese are equal

On Apr 30, 2024


C
KP

Answered

Some researchers have conjectured that stem-pitting disease in peach-tree seedlings might be controlled with weed and soil treatment.An experiment was conducted to compare peach-tree seedling growth with soil and weeds treated with one of two herbicides.In a field containing twelve seedlings,six were randomly selected throughout the field and assigned to receive Herbicide A.The remaining six were to receive Herbicide B.Soil and weeds for each seedling were treated with the appropriate herbicide,and at the end of the study period the height in centimeters was recorded for each seedling.The following results were obtained. Some researchers have conjectured that stem-pitting disease in peach-tree seedlings might be controlled with weed and soil treatment.An experiment was conducted to compare peach-tree seedling growth with soil and weeds treated with one of two herbicides.In a field containing twelve seedlings,six were randomly selected throughout the field and assigned to receive Herbicide A.The remaining six were to receive Herbicide B.Soil and weeds for each seedling were treated with the appropriate herbicide,and at the end of the study period the height in centimeters was recorded for each seedling.The following results were obtained.   Let W be the Wilcoxon rank sum test statistic equal to the sum of the ranks assigned to the Herbicide A group.What is the expected value of W under the null hypothesis that the two distributions are the same? A) 36 B) 39 C) 54 D) 78 Let W be the Wilcoxon rank sum test statistic equal to the sum of the ranks assigned to the Herbicide A group.What is the expected value of W under the null hypothesis that the two distributions are the same?

A) 36
B) 39
C) 54
D) 78

On Apr 29, 2024


B