Exam 11: Analysis of Variance

Sound engineers studied factors that might affect the output (in decibels) of a rock concert speaker system. The results of their ANOVA tests are shown (some information is missing). Source of Variation SS df MS F P -value Amplifier 99.02344 99.02344 0.005718 Position 93.98698 31.32899 3.215807 0.051003 Interaction 10.15365 3 3.384549 0.347412 0.791505 Error 155.875 16 9.742188 Total 359.0391 23 Which is the number of amplifiers and positions tested?

For this one-factor ANOVA (some information is missing), what is the F-test statistic? Source Sum of Squares df Mean Square F Treatment 654 218 Error 3,456 128 Total 4,110

ANOVA is a procedure intended to compare the variances of several groups (treatments).

Tukey's test pools all the sample variances.

A multinational firm manufactures several types of 1280 × 1024 LCD displays in several locations. They designed a sampling experiment to analyze the number of pixels per screen that have significant color degradation after 52,560 hours (six years of continuous use) using accelerated life testing. The Excel ANOVA table for their experiment is shown below. Some table entries have been obscured. The response variable (Y) is the number of degraded pixels in a given display. Source of Variation SS df MS F Country of origin 202.9 101.45 4.163475 Display type 233.2333 58.30833 Interaction 147.7667 18.47084 Error 1096.5 45 24.36667 Total 1680.4 59 How many countries were studied?

A firm is concerned with variability in hourly output at several factories and shifts. Here are the results of an ANOVA using output per hour as the dependent variable (some information is missing). Source Sum of Squares df Mean Square FRatio P-value Factory 19012.5 1 19012.5 26.427 0.000 Shift 258.333 2 129.167 0.180 0.838 Factory*Shift 80908.333 2 40454.167 56.230 Error 8633.333 12 719.444 Total 108812.5 17 6400.735 At ? = 0.01 the effect of factory is:

Degrees of freedom for the between-group variation in a one-factor ANOVA with n₁ = 5, n₂ = 6, n₃ = 7 would be:

Tukey's test for five groups would require 10 comparisons of means.

A firm is concerned with variability in hourly output at several factories and shifts. Here are the results of an ANOVA using output per hour as the dependent variable (some information is missing). Source Sum of Squares df Mean Square FRatio Factory 19012.5 1 19012.5 26.427 Shift 258.333 2 129.167 0.180 Factory*Shift 80908.333 2 40454.167 56.230 Error 8633.333 12 719.444 Total 108812.5 17 6400.735 The number of observations in each treatment cell (row-column intersection) is:

Here is an Excel ANOVA table that summarizes the results of an experiment to assess the effects of ambient noise level and plant location on worker productivity. The test used ? = 0.05. Source of Variation SS df MS F P-value F crit Plant location 3.0075 3 1.0025 2.561 0.1199 3.862 Noise level 8.4075 3 2.8025 7.160 0.0093 3.863 Error 3.5225 9 0.3914 Total 14.9375 The experimental design and ANOVA appear to be:

Refer to the following partial ANOVA results from Excel (some information is missing). The response variable was Y = maximum amount of water pumped from wells (gallons per minute). Source of Variation SS df MS F Depth of well 2450 2 1225 Age of well 364.667 3.8371 Interaction 32.667 4 0.1719 Error 855.333 18 47.519 Total 3702.667 26 The MS for age of well is:

Sample sizes must be equal in one-factor ANOVA.

Systolic blood pressure of randomly selected HMO patients was recorded on a particular Wednesday, with the results shown here. An ANOVA test was performed using these data. Patient Age Group Under 20 20 to 29 30 to 49 50 and Over 105 110 122 139 113 101 114 115 108 112 128 136 114 127 124 124 123 123 125 123 Degrees of freedom for the between-treatments sum of squares would be:

A firm is studying the effect of work shift and parts supplier on its defect rate (dependent variable is defects per 1000). The resulting ANOVA results are shown below (some information is missing). Source Sum of Squares df Mean Square F P-Value Shift 704.07 2 352.04 11.93 0.001 Supplier 19.6 9.8 0.33 0.72 Shift*Supplier 430.75 4 107.69 3.65 0.014 Error 1062.05 36 29.5 Total 2216.47 44 How many suppliers were there?

To test the null hypothesis H₀: μ₁ = μ₂ = μ₃ using samples from normal populations with unknown but equal variances, we:

Degrees of freedom for the between-group variation in a one-factor ANOVA with n₁ = 8, n₂ = 5, n₃ = 7, n₄ = 9 would be:

Analysis of variance is a technique used to test for:

Hartley's test is the largest sample mean divided by the smallest sample mean.

Which Excel function gives the right-tail p-value for an ANOVA test with a test statistic F_calc = 4.52, n = 29 observations, and c = 4 groups?

A multinational firm manufactures several types of 1280 × 1024 LCD displays in several locations. They designed a sampling experiment to analyze the number of pixels per screen that have significant color degradation after 52,560 hours (six years of continuous use) using accelerated life testing. The Excel ANOVA table for their experiment is shown below. Some table entries have been obscured. The response variable (Y) is the number of degraded pixels in a given display. Source of Variation SS df MS F P -value F crit Country of origin 202.9 101.45 4.163475 0.021927 Display type 233.2333 58.30833 Interaction 147.7667 18.47084 Error 1096.5 45 24.36667 Total 1680.4 59 The F statistic for display effect is: