Question

Researchers are studying the effect of tilling(no-till vs. traditional tilling) on the production of corn yield . They formulated the following research question. Does yield tend to be lower on no-tilled plots than on traditionally-tilled plots?

In the study, they divided up fields on 20 different farms into equal sized plots. One plot on each farm was farmed with conventional tilling methods for 5 years, whereas the other plot was farmed with not-till methods for the same five year period. In the final year, they measured the yield (tons/acre) of corn on each of the 40 plots. Assume that population differences are normally distributed.

Here is the R output:

Paired t-test

data: notill and traditional

t = 2.3803, df = 19, p-value = 0.02793

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

1.966947 30.633053

sample estimates:

mean of the differences

16.3

part a) Test the research question at the 5% level of significance.

part b) State and interpret the 95% confidence interval for the difference in population means.

Answer 1

We have porvided R output t = 3.3803, df = 19 , pvalue = v0.02791 alpha =- 0.05 H_0: mud = 0 v & Ha: mu_d noequalto 0 (two tailed) sample astimate d^- =16.3 mean difference df n - 1 = 13 n = 20 standard deviation of difference is not provi ded) Test statistic t_slat = d^-/Sd/squareroot n = 16.3/Sd/squareroot 20 Pvalue = 2 times P(t_0.05,10 lessthan \2 38031\ = 0.027793 Reject Ho if t_slat greaterthan t_critical orPvalue lessthanorequalto level of significance = alpha 95% confidence interval given (1. 9669447,30.633053) interpretation weare 95% confident that the different integral population lies between (1. 9669447,30.633053) confidence interval formula Lower bound d^-- - tg/2 times Sd/squareroot n upper bound d^-- -+tg/2 times Sd/squareroot n t_alpha/2 = t_critical at alpha = 0.05 df=19 t_critical = plusminus 2.093024 (two talled) we can find Standard deviation using test statistic t_slat = d^-/Sd/squareroot n 2.3803 = 16.3/Sd/squareroot 20 Sd = 16.3 times

Note :

Standard deviation is not provided

But we can find using given output

Test statistic , mean difference and sample is given

So standard deviation is 30.6246

By using this standard deviation all answers are coming true.

Confidence interval is also correct by putting SD= 30.6246.