Better barley Does drying barley seeds in a kiln increase the yield of barley? A famous experiment by William S. Gosset (who discovered the t distributions) investigated this question. Eleven pairs of adjacent plots were marked out in a large field. For each pair, regular barley seeds were planted in one plot and kiln-dried seeds were planted in the other. A coin flip was used to determine which plot in each pair got the regular barley seed and which got the kiln-dried seed. The following table displays the data on barley yield (pound per acre) for each plot.

Do these data provide convincing evidence at the $\alpha =0.05$level that drying barley seeds in a kiln increases the yield of barley, on average?

We were told that on a big field, eleven pairs of neighboring plots were marked out, with ordinary barley seeds planted in one plot and kiln-dried seeds placed in the other.

We need to find out that do these data provide convincing evidence at the $\alpha =0.05$level that drying barley seeds in a kiln increases the yield of barley, on average

Given:

$$\begin{gathered} n=Samplesize=11 \\ \alpha =Significancelevel=0.05 \end{gathered}$$

Let us determine the difference between regular barley seeds and kiln-dried seeds

Now we will determine the mean of values of the difference:

$$\begin{gathered} \overline{x}=\frac{-106+20-101+33-72-62+36-38+70-127-24}{11} \\ =\frac{-371}{11}\approx -33.7273 \end{gathered}$$

Now we will determine the standard deviation

$s=\sqrt{\frac{{\displaystyle \sum _{n=1}^{10}{\left(Difference-\overline{x}\right)}^2}}{n-1}}$

$$\begin{gathered} s=\sqrt{\frac{{\left(106+33.7273\right)}^2+{\left(20+7273\right)}^2+{\left(-101+33.7273\right)}^2+{\left(33+33.7273\right)}^2 \\ +{\left(-72+33.7273\right)}^2+{\left(-62+33.7273\right)}^2+{\left(36+33.7273\right)}^2+{\left(-38+33.7273\right)}^2 \\ +{\left(\left(70+33.7273\right)\right)}^2+{\left(-127+3.7273\right)}^2+{\left(-24+33.7273\right)}^2}{11-1}} \\ \approx 66.1711 \end{gathered}$$

Now we will carry out a hypothesis test for the population mean difference.

Here we have:

$$\begin{gathered} Populationmeandifference={\mu }_d \\ {H}_0=Nullhyphothesis \\ {H}_a=Alternativehyphothesis \\ {H}_0:{\mu }_d=0 \\ {H}_a:{\mu }_d\ne 0 \\ Nowwewilldeterminethevalueofteststatistics: \\ t=\frac{\overline{x}-{\mu }_d}{{\displaystyle \raisebox{1ex}{$s$}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}} \end{gathered}$$

$t=\frac{-33.7273-0}{{\displaystyle \raisebox{1ex}{$66.1711$}\!\left/ \!\raisebox{-1ex}{$\sqrt{11}$}\right.}}\approx -1.690$

The P-value is the probability of obtaining the value of test statistics.

Degree of freedom $=11-1=10$

The test is a two-tailed test so we double the boundaries of the P-value.

$$\begin{gathered} 0.05<P<0.10 \\ nowatt=-1.690andDegreeoffreedom=10,Pvalue=0.06095 \end{gathered}$$

We have to reject the null hypothesis if the probability value is less than the hypothesis value.

$P>0.05\Rightarrow failtoreject{H}_0$

This demonstrates that there is no convincing evidence that kiln-drying barley seeds increases barley yields on average.