Some researchers have conjectured that stem-pitting disease in peach-tree seedlings might be controlled with weed and soil treat
ment. An experiment is conducted to compare peach-tree seedling growth when the soil and weeds are treated with one of two herbicides. In a field containing 20 seedlings, 10 are randomly selected throughout the field and assigned to receive herbicide A. The remainder of the seedlings is assigned to receive herbicide B. Soil and weeds for each seedling are treated with the appropriate herbicide, and at the end of the study period, the height in centimeters is recorded for each seedling. The following results are obtained: Herbicide A = 94.5 cm s1 = 10 cm Herbicide B = 109.1 cm s2 = 9 cm Required:
1. What is a 90% confidence interval (use the conservative value for the degrees of freedom) for μ2 – μ1? a. 14.6 ± 7.00 b. 14.6 ± 7.38 c. 14.6 ± 7.80 d. 14.6 ± 9.62
2. Suppose we wish to determine if there tends to be a difference in height for the seedlings treated with the different herbicides. To answer this question, we decide to test the hypotheses H0: μ2 – μ1 = 0, Ha: μ2 – μ1 ≠0. What is the value of the two-sample t statistic?
a. 14.6
b. 3.43
c. 2.54
d. 7.80
3. What can we say about the value of the P-value?
a. P-value < 0.01
b. 0.01 < P-value < 0.05
c. 0.05 < P-value < 0.10
d. P-value > 0.10
4. If you use the 0.05 level of significance, what conclusion would you reach?
a. The evidence is not sufficiently strong to reject the null hypothesis.
b. There is sufficient evidence to reject the null hypothesis.
c. Cannot be determined by the information given.
4. b. There is sufficient evidence to reject the null hypothesis.
Step-by-step explanation:
The hypothesis test is used to test the null hypothesis against the alternative hypothesis. The p value of the test determines whether to accept or reject the null hypothesis. In the given scenario the p value is less than 0.05 so there is sufficient evidence against the null hypothesis in favor of alternative hypothesis.
