PLease show step and calculation formula that involed A plan
PLease show step and calculation (formula that involed)
A plant biologist studied whether physical stress can slow the growth of soybean plants. Prior to the study, a hypothesis was formulated, based on other plant species, that physical stress should make growth slower. Two groups of 13 plants were grown under identical conditions in a greenhouse. The treatment group was shaken by a mechanical device for 20 minutes twice every day (simulating stress caused by winds), while the control group was not shaken. After 16 days of growth, plant height in cm was measured, and the following data were obtained. Using the following summary data, do a hypothesis test with = 0.05 that addresses the plant biologist’s prior hypothesis, and state the conclusion of the test.
Control Treatment
n 13 13
Sample Mean 30.59 27.78
Sample SD 2.13 1.73
Solution
Let mu1 be the mean for control group and mu2 treatment groups
H0: mu1 = mu2
Ha: mu1 > mu2
Right tailed test
As sample size <30, t test can be used.
standard error of difference = 2.744
The mean of Group One minus Group Two equals 2.8100
t = 1.0240
df = 24
The two-tailed P value equals 0.3160
By conventional criteria, this difference is considered to be not statistically significant.
Hence at 0.05 level there is no evidence to support biologist claim
| Group | Group One | Group Two |
|---|---|---|
| Mean | 30.5900 | 27.7800 |
| SD | 7.6798 | 6.2376 |
| SEM | 2.1300 | 1.7300 |
| N | 13 | 13 |
