Question

For this problem, five cotton varieties were selected from a larger study of yield stability. The objective was to determine if there are differences among the mean weight of the fiber for these varieties and if so, in the next homework problem to determine those differences. For each variety, data were collected at harvest from four plots of each variety in the yield stability test. The researcher recorded the data in units of “thousands of fibers per seed.” There were no missing data.

1. (a) Identify the populations and random variables of interest.

2. (b) Give a model for this study (i.e, the distributions of the random variables and all other statistical

   assumptions).

3. (c) Set up an appropriate initial set of hypothesis to be tested. Give the test statistic and its

   distribution. Using " = 0.01, give the critical region for the test.

4. (d) SAS® was used to carry out an analysis of variance. From the computer output,

   SSVarieties = 57.0636 and SSError = 23.0537.

   Use this information to test your initial hypothesis using the test that you set up in part (c). Summarize your calculations in analysis of variance (ANOVA) table. Your conclusion must be stated in the context of the problem.

Answer 1

(a)

Population - Cotton Varieties in the plots

Random variables of interest - Mean weight of the fiber for Cotton Varieties

(b)

We will conduct One Way Anova test for the study, The statistical assumptions of the Anova test are
(i) All populations involved follow a normal distribution.
(ii) All populations have the same variance (or standard deviation).
(iii) The samples are randomly selected and independent of one another.

(c)

Null Hypothesis H0: Mean weight of the fiber for all five cotton varieties are equal.

Alternative Hypothesis Ha: At least one of the mean weight of the fiber of any cotton variety is different than the other.

Between degree of freedom = Number of cotton varieties - 1 = 5 -1 = 4

Within degree of freedom = Number of observations - Number of cotton varieties = 5 * 4 - 5 = 15

The distribution of test statistic follow F distribution with numerator df = 4 and denominator df = 15

For = 0.01, Critical value of F at df = 4, 15 is 4.89

We reject the null hypothesis H0 if the test statistic F > 4.89

(d)

MSVarieties = SSVarieties / Between df = 57.0636 / 4 = 14.2659

MSError = SSError / Within df = 23.0537 / 15 = 1.536913

Test statistiic, F = MSVarieties / MSError

= 14.2659 / 1.536913

= 9.28

Since the observed F (9.28) is greater than the critical value, we reject the null hypothesis H0 and conclude that there is significant evidence to reject the claim that mean weight of the fiber for all five cotton varieties are equal.

Anova Table -

SSTotal = SSVarieties + SSError = 57.0636 + 23.0537 = 80.1173

Source	DF	SS	MS	F
Varieties	4	57.0636	14.2659	9.28
Error	15	23.0537	1.536913
Total	19	80.1173