Question

ff 1 An agricultural researcher reasons as follows: A heavy application of potassium fertilizer to the grasslands in the spring seems to cause lush early growth but depletes the potassium before the growing season ends. So spreading the same amount of potassium over the growing season might increase yields. He therefore compares two treatments: 100 pounds per acre of potassium in the spring (Treatment 1) and 50. 25 and 25 pounds per acre in the spring, early summer, and late summer (Treatment 2). The experiment is continued over several years because growing conditions may vary from year to year. The table below gives the yields (in pounds of dry matter per acre). To be certain that growing conditions are the same, each treatment is applied over one half of each field. applied Treatment 1 Treatment 2 Year Year 1 3902 3970 Year 2 4281 4271 Year 3 5135 5440 Year 4 5350 5490 Year 5 5746 6028 etc. etc. etc. Test whether Treatment 2 leads to higher average yields than Treatment 1. a State HO and H1. Your MUST show the five-step process for determining HO and H1 (1 pt.)

Answer 1

a) We have to test :

$H_0 : \mu_1 = \mu_2 \ \ against \ \ H_1 : \mu _2>\mu_1 \ \ \\ alternatively\ \ \\ H_0: \mu_d = 0 \ \ against H_1: \mu_d<0$

$\mu_d = \mu_1-\mu_2$

b) Here we conduct a paired t test .As the data is of same measure using different aplliance .[ eg, testing an effect of something , data is of with or without the effect] .

Test statistic :

$T = \frac{\bar{d}}{\sqrt{s^2/n}}$ ;  

where

$\bar{d}$ = average difference  

s² = sample variance of d  

n = sample size.

Critical region : Reject H₀ if observed T < $t_{1-\alpha,4}$

where alpha is the significance level .

Now, T = -2.59  

and at 95% confidence level  $t_{0.95,4} = -2.13$

Note that T< t

c) SO, we have enough evidence to reject null hypothesis at 95% confidence level .

i.e treatment 2 leads to higher average than trt 1