The following data are from an experiment to look at weight gain [in gms] of lab mice under 3 different


  1. The following data are from an experiment to look at weight gain [in gms] of lab

mice under 3 different diets which have low, medium, and high amounts of protein and other nutrients. It was also felt that male mice might respond to the diets differently than female mice, so sex of the animal was also noted.

FEMALES MALES

(cell means) (cell means) [row means, sexes combined]

LOW 40.8 49.2

40.5 (41.73) 43.8 (43.53) [42.63]

43.9 37.6

MED 51.5 50.9

50.0 (51.13) 57.9 (56.23) [53.68]

51.9 59.9

HIGH 62.7 74.0

56.4 (61.27) 72.1 (73.33) [67.30]

64.7 73.9

MEAN 51.38 57.70

  1. At the .10 level of significance, test for the presence of interaction. Use an interaction plot of the means to interpret the result.
  2. At the .10 level of significance, test for the overall (main) effect of sex on the mean weight gain. Refer to the interaction plot in your interpretation of the test result.
  3. At the .05 level of significance, test for the overall effect of the three diets on the mean weight gain. Again, using the interaction plot, interpret the result of the statistical test.
  4. Compute a 90% confidence interval for the average weight gain on the LOW diet.
  5. Compute a 90% confidence interval for the difference between the male and female average weight gain for the HIGH diet.
Total 55795.990 18
Corrected Total 2255.163 17
  1. R Squared = .929 (Adjusted R Squared = .899)

2. The feather density of 4 species of penguin was compared with ANOVA and found to

be different. After completing the ANOVA display table, use Tukey's HSD to analyze the data to find out where the differences in feather density occur (use alpha = .05). Summarize your findings by writing the means down in order and using the "underline" technique to display which means can and cannot be separated.

SPECIES

Magellanic Adelie Gentoo Macaroni

n 5 5 5 5

Mean 53.533 83.200 78.722 74.667

SOURCE OF VARIATION df SS MS F P-value

Total(corrected) 3125.880

Between species 2588.212 25.73

Within species (error) 537.668 33.604

3. Five hospitals were surveyed regarding a specific surgical procedure designed to

improve functioning of certain joints impaired by disease. In this study, "no improvement," and "improvement" were specifically defined and carefully measured. Interest lay in finding any differences in success between the hospitals.

Hospitals

A B C D E Total

no improvement 13 5 8 21 43 90

improvement 16 16 35 51 10 128

Total 29 21 43 72 53 218

  1. If the five hospitals had the same distribution of successes, compute and display the table of expected counts for hospital C look like (that is to say, under the null hypothesis, what are the expected frequencies for hospital C)?
  2. At the .01 level of significance test the null hypothesis that the five hospitals have the same distribution of success in the surgical procedure performed.
  3. [No testing here, just data inspection] The researcher suspected that the five hospitals can be combined (some of the hospitals have similar outcomes). Compute the column percentages so that for every hospital, one can see what percentage of patients fall in each of the two success categories. Based on that information ALONE, indicate which hospitals seem to "go together" with which other hospitals.

4. An experiment was implemented to study the effect of row spacing (0.25, 0.5 and 1 m)–on the moisture content of soils in which a particular plant variety was grown under a specific type of irrigation practice. Three separate, fairly uniform areas were found that could hold three plots each, for a total of 9 plots. The three spacing levels were randomized so that each plot group had one plot of each spacing.

Treatment Spacing Level Plot 1 Plot 2 Plot 3

1 0.25 30.7 24.7 16.9

2 0.50 22.7 15.0 13.6

3 1.00 23.0 17.1 15.2

  1. Write the null and alternate hypotheses (in model equation form) for the ANOVA test,
    explaining all terms.
  2. Show the ANOVA table and draw a conclusion from your hypothesis test conducted at the 0.10 significance level.
  3. Calculate the estimated standard error for any single treatment group mean.
  4. Calculate the standard error of the difference between any two observed treatment means.

5. A completely randomized design was used to study the effects of five different soil

amendments on growth of a particular plant species. Each treatment group, including a no amendment "control" group was replicated three times for a total N = 18 observations. The treatments were defined as follows:

A: compost + Nitrogen

B: compost + Nitrogen + Potassium

C: peat moss + Phosphorus

D: peat moss + Nitrogen + Calcium

E: peat moss + Calcium + Potassium

F: no amendments (the control)

You wish to make the following comparisons using contrasts:

  1. the effect of Nitrogen vs. no Nitrogen
  2. the effect of compost vs. peat moss
  3. the effect of fertilizer "blends" vs. single nutrient additions
  1. Derive numeric contrast coefficients for each treatment group to estimate the contrast values.
  2. Which of the above contrasts are orthogonal to each other?
  3. If the MSE for this study turned out to be 11, compute the standard error for contrast i)

Nitrogen vs. no Nitrogen.

Price: $24.8
Solution: The downloadable solution consists of 11 pages, 1380 words and 1 charts.
Deliverable: Word Document


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