Question

A consumer advocate examines whether the longevity of car batteries (measured in years) is affected by...

A consumer advocate examines whether the longevity of car batteries (measured in years) is affected by the brand name (factor A) and whether or not the car is kept in a garage (factor B). Interaction is suspected. The results are shown in the accompanying table.

Brand Name of Battery
Kept in Garage? A B C
Yes 8, 7, 7 7, 6, 7 8, 9, 10
No 6, 7, 6 5, 6, 4 5, 7, 7

Click here for the Excel Data File

a-1. Construct an ANOVA table. (Round intermediate calculations to at least 4 decimal places. Round "SS", "MS" to 4 decimal places and "F", "p-value" to 3 decimal places.)

a-2. At the 5% significance level, Is there interaction between the brand name and whether a car is garaged?

  • Yes since the p-value for interaction is less than the significance level.
  • Yes since the p-value for interaction is greater than the significance level.
  • No since the p-value for interaction is greater than the significance level.
  • No since the p-value for interaction is less than the significance level.


b. At the 5% significance level, can you conclude that the average battery lives differ by brand name?


  • No since the p-value for brand name is less than the significance level.
  • Yes since the p-value for brand name is less than the significance level.
  • No since the p-value for brand name is less than the significance level.
  • No since the p-value for brand name is greater than the significance level.


c. At the 5% significance level, can you conclude that the average battery lives differ depending on whether a car is garaged?


  • Yes since the p-value for garage is greater than the significance level.
  • Yes since the p-value for garage is less than the significance level.
  • No since the p-value for garage is less than the significance level.
  • No since the p-value for garage is less than the significance level.

Homework Answers

Answer #1
Source of variation Degree of Freedom Sum of Squares Mean Sum of Squares F Value
Between levels of Factor A p-1
Between levels of Factor B q-1
Interaction between A and B (p-1)(q-1)
Error pq(m-1)
Total mpq-1

Above is the ANOVA table for 2 way analysis of data with m observation per cell.

Where denotes mean of row, denotes mean of column, denotes mean of row and column, refers to mean of all observation and i = 1, 2, 3, ...,p & j = 1, 2, 3, ...., q

In our case, Factor B has 2 levels, ie p = 2 and factor A has 3 levels, q = 3.

Calculation,

required values:

1 2 3
1 7.3333 6.6667 9 7.6667
2 6.3333 5 6.3333 5.8889
6.8333 5.8333 7.6667

= 6.778

a1) ANOVA Table

Source of variation Degree of Freedom Sum of Squares Mean Sum of Squares F Value
Between levels of Factor A 1 14.222 19.692
Between levels of Factor B 2 5.0555 6.991
Interaction between A and B 2 1.0555 1.462
Error 12 0.7222
Total 17

a2) No, since the p-value for interaction (0.730) is more than the significance level (0.05)

b) Yes, since the p-value for brand name (0.004) is less than the significance level 0.05.

c) Yes, since the p-value for garage (0.000) is less than the significance level 0.05.

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A consumer advocate examines whether the longevity of car batteries (measured in years) is affected by...
A consumer advocate examines whether the longevity of car batteries (measured in years) is affected by the brand name (factor A) and whether or not the car is kept in a garage (factor B). Interaction is suspected. The results are shown in the accompanying table. Garage Brand Years Yes A 6 Yes A 8 Yes A 7 No A 6 No A 5 No A 5 Yes B 6 Yes B 6 Yes B 6 No B 5 No B...
A researcher conducts a two-way ANOVA test with interaction and provides the following ANOVA table. a....
A researcher conducts a two-way ANOVA test with interaction and provides the following ANOVA table. a. Find the missing values in the ANOVA table. (Round your answers to 2 decimal places.)   ANOVA   Source of Variation SS df        MS       F p-value F crit   Sample 766.98   2 MSB =    Ffactor B =    0.0109   3.885    Columns 12,122.69   1 MSA =    Ffactor A =    0 4.747    Interaction 57.60   2 MSAB =    FInteraction =    0.6148 3.885    Within 682.14 12 MSE =       Total 13,629.41 17 b....
A management consultant wants to determine whether the age and gender of a restaurant’s wait staff...
A management consultant wants to determine whether the age and gender of a restaurant’s wait staff influence the size of the tip the customer leaves. Three age brackets (factor A in columns: young, middle-age, older) and gender (factor B in rows: male, female) are used to construct a two-way ANOVA experiment with interaction. For each combination, the percentage of the total bill left as a tip for 10 wait staff is examined. The following ANOVA table is produced. ANOVA Source...
Consider the following hypotheses: H0: μ ≥ 208 HA: μ < 208 A sample of 80...
Consider the following hypotheses: H0: μ ≥ 208 HA: μ < 208 A sample of 80 observations results in a sample mean of 200. The population standard deviation is known to be 30. (You may find it useful to reference the appropriate table: z table or t table) a-1. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal...
A multinomial experiment produced the following results: (You may find it useful to reference the appropriate...
A multinomial experiment produced the following results: (You may find it useful to reference the appropriate table: chi-square table or F table) Category 1 2 3 4 5 Frequency 57 63 70 55 55 a. Choose the appropriate alternative hypothesis to test if the population proportions differ. All population proportions differ from 0.20. Not all population proportions are equal to 0.20. b. Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final...
A two-way analysis of variance experiment with no interaction is conducted. Factor A has three levels...
A two-way analysis of variance experiment with no interaction is conducted. Factor A has three levels (columns) and Factor B has seven levels (rows). The results include the following sum of squares terms: SST = 346.9 SSA = 196.3 SSE = 79.0 a. Construct an ANOVA table. (Round intermediate calculations to at least 4 decimal places. Round "SS" to 2 decimal places, "MS" to 4 decimal places, "F" to 3 decimal places.) Source SS df MS F p-value Rows Columns...
You suspect that an unscrupulous employee at a casino has tampered with a die; that is,...
You suspect that an unscrupulous employee at a casino has tampered with a die; that is, he is using a loaded die. In order to test this claim, you roll the die 282 times and obtain the following frequencies: (You may find it useful to reference the appropriate table: chi-square table or F table)        Category   1   2   3   4   5   6 Frequency   64   57   46   38   38   39 Click here for the Excel Data File a. Choose the...
Consider the following competing hypotheses and accompanying sample data. (You may find it useful to reference...
Consider the following competing hypotheses and accompanying sample data. (You may find it useful to reference the appropriate table: z table or t table) H0: p1 − p2 ≥ 0 HA: p1 − p2 < 0 x1 = 250 x2 = 275 n1 = 400 n2 = 400 a. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal...
Three different brands of car batteries, each having a 42 month warranty, were included in a...
Three different brands of car batteries, each having a 42 month warranty, were included in a study of battery lifetime. A random sample of batteries of each brand was selected, and lifetime, in months, was determined. An analysis of variance was conducted and the p-value was 0.053. The hypothesis that all mean lifetimes are the same was retained at the 5% level. Multiple comparisons of the means were also carried out with the following (somewhat contradictory) findings: The Tukey's 95%...
Consider the following competing hypotheses and accompanying sample data drawn independently from normally distributed populations. (You...
Consider the following competing hypotheses and accompanying sample data drawn independently from normally distributed populations. (You may find it useful to reference the appropriate table: z table or t table) H0: μ1 − μ2 = 0 HA: μ1 − μ2 ≠ 0 x−1x−1 = 75 x−2x−2 = 79 σ1 = 11.10 σ2 = 1.67 n1 = 20 n2 = 20 a-1. Calculate the value of the test statistic. (Negative values should be indicated by a minus sign. Round all intermediate...