Viruses are infectious agents that often cause diseases in plants. Different viruses have different potency levels, and this fact can be used to detect whether a new virus is infecting plants in the field. In a potency comparison experiment, two viruses were placed on a tobacco leaf of 10 randomly selected tobacco plants in a field. The viruses were randomly assigned to one-half of each of the leaves. The table below presents the potency of the viruses, as measured by the number of lesions appearing on the leaf half.

Question

Viruses are infectious agents that often cause diseases in plants. Different viruses have different potency levels, and this fact can be used to detect whether a new virus is infecting plants in the field. In a potency comparison experiment, two viruses were placed on a tobacco leaf of 10 randomly selected tobacco plants in a field. The viruses were randomly assigned to one-half of each of the leaves. The table below presents the potency of the viruses, as measured by the number of lesions appearing on the leaf half.

Leaf # 1 2 3 4 5 6 7 8 9 10
Virus X 9 8 3 4 8 4 17 3 14 20
Virus Y 19 8 13 5 16 8 17 6 19 17

Test the hypothesis that there is no difference in mean number of lesions for Virus X and Virus Y.

Answer 1

To test the hypothesis that there is no difference in the mean number of lesions for Virus X and Virus Y, we can use a statistical test called the independent samples t-test. This test compares the means of two independent groups to determine if they are significantly different from each other.

Here's how you can perform the independent samples t-test for this experiment:

1. Define the null and alternative hypotheses:
- Null hypothesis (H0): There is no difference in the mean number of lesions for Virus X and Virus Y. μx = μy.
- Alternative hypothesis (Ha): There is a difference in the mean number of lesions for Virus X and Virus Y. μx ≠ μy.

2. Calculate the mean and standard deviation for each group (Virus X and Virus Y) using the given data.

For Virus X:
- Mean (x̄x) = (9 + 8 + 3 + 4 + 8 + 4 + 17 + 3 + 14 + 20) / 10
- Standard deviation (σx) = Calculate the sample standard deviation using the formula.

For Virus Y:
- Mean (x̄y) = (19 + 8 + 13 + 5 + 16 + 8 + 17 + 6 + 19 + 17) / 10
- Standard deviation (σy) = Calculate the sample standard deviation using the formula.

3. Calculate the t-value using the formula:

t = (x̄x - x̄y) / sqrt((σx^2 / n) + (σy^2 / n))

where:
x̄x and x̄y are the means of Virus X and Virus Y, respectively,
σx and σy are the standard deviations of Virus X and Virus Y, respectively,
n is the number of samples (in this case, n = 10).

4. Determine the degrees of freedom (df) for the t-distribution. In this case, df = n1 + n2 – 2, where n1 and n2 are the sample sizes of Virus X and Virus Y, respectively.

5. Look up the critical t-value for the desired significance level (e.g., α = 0.05) and the degrees of freedom obtained in step 4.

6. Compare the calculated t-value from step 3 with the critical t-value from step 5. If the calculated t-value is greater than the critical t-value (in absolute value), then we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

Performing these steps should yield the final result of whether there is a significant difference in the mean number of lesions for Virus X and Virus Y.