Question

When a virus is placed on an oak leaf, small lesions appear on the leaf. Researchers believe different strains of the virus will produce different effects. They wished to test this idea by comparing the mean number of lesions produced by two different strains of the virus. In the experiment, Strain 1 (VS1) was applied to 10 oak leaves and Strain 2 (VS2) was applied to another 10 oak leaves. The lesions that appear on each leaf were then counted. The raw data are shown below:

Strain 1 18 21 17 22 20 20 20 18 19 17

Strain 2 11 18 10 14 17 14 10 14 17 11

a. Symbolically state the null and alternate hypotheses.  

b. Assuming the above data were normally distributed, what is the appropriate test for this experiment?

i. F-test ii. Independent t-test iii. Z-test iv. F-test then paired t-test v. F-test then independent t-test vi. F-test then z-test vii. Paired t-test

c. What is the mean number of lesions for each group?

d. Symbolically state the null and alternate hypotheses for the F-test (use the appropriate symbols)

e. If the variance for VS1 is 2.84 and the variance for VS2 is 9.15, calculate the F statistic.  

f. Given α = 0.05, state the critical value for the F statistic and whether you would accept or reject the F-test null-hypothesis?

g. Regarding the ensuing t-test, is this 1- or 2-tails and what direction is it if 1- tailed?  

h. If the pooled variance (s2 p) is 6.0, calculate the test statistic for the t-test.

i. What are the degrees of freedom for this test? Given α = 0.05, would you accept or reject the t-test null hypothesis? Why?

Answer 1

a.

where are the true mean number of lesions produced by strains 1 and strains 2 respectively.

b.

Since oak leaves used for both strains were different and independent, and we do not know the population standard deviations of of lesions produced by strains 1 and strains 2, we use F test to compare the variances of two groups and based on that we use independent t-test using pooled variance or welch t test.

v. F-test then independent t-test

c.

= (18 + 21 + 17 + 22 + 20 + 20 + 20 + 18 + 19 + 17)/10 = 19.2

= (11 + 18 + 10 + 14 + 17 + 14 + 10 + 14 + 17 + 11) / 10 = 13.6

d.

e.

F statistic = VS2 / VS1 = 9.15 / 2.84 = 3.22

f.

Numerator df = n2 - 1 = 10 - 1 = 9

Denominator df = n1 - 1 = 10 - 1 = 9

Critical value for the F statistic at α = 0.05 and df = 9, 9 is 4.03

Since the observed F statistic is less than the critical value, we accept the F-test null-hypothesis and conclude that there is no significant evidence that variances of the two groups are not equal.

g.

Since the alternative hypothesis in part (a) is non-directional (unequal sign), this is a two tail test.

h.

Standard error, SE =

= 1.095445

Test statistic for the t-test, t = ( - ) / SE

= (19.2 - 13.6) / 1.095445

= 5.112

i.

Degree of freedom = n1 + n2 - 2 = 10 + 10 - 2 = 18

Critical value of t at α = 0.05 for two tail test and df = 18 is,  2.1

We reject null hypothesis if t < -2.1 or t > 2.1

Since the observed test statistic lies in the rejection region, we reject the t-test null hypothesis and conclude that there is significant evidence that the true mean number of lesions produced by strains 1 and strains 2 are different.