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. The following table displays the data on yield (lb/acre).

(a) How can the Random condition be satisfied in this study?

(b) Perform an appropriate test to help answer the research question. Assume that the Random condition is met. What conclusion would you draw?

Random sampling provides us with non-aligned data from the population. When samples are collected in an erroneous manner, the data is prone to intolerance.

You will randomly assign one plot to be planted with conventional barley seeds, while the next plot will be planted with kiln-dried seeds in each pair of adjacent plots.

Determine the difference between "Kiln" and "Regular" for each pair.

The mean is the sum of all values divided by the number of values:

$\overline{x}=\frac{106-20+\dots +127+24}{11}$

$\approx 33.7273$

$n$is the number of values in the data set.

The variance is the sum of squared deviations from the mean divided by $n-1$:

localid="1650366695337" $s^2=\frac{(106-33.7273{)}^2+\dots .+(24-33.7273{)}^2}{11-1}$

$\approx 3980.5620$

The standard deviation is the square root of the variance:

localid="1650366712856" $s=\sqrt{3980.5620}$

$\approx 66.1711$

Determine the hypotheses:

$H_0:\mu =0$

$H_a:\mu >0$

Determine the value of the test statistic:

localid="1650366606373" $t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}$

$=\frac{33.7273-0}{66.1711/\sqrt{11}}$

$=1.690$

The $P-$value is the chance of getting the test statistic's result, or a number that is more severe. The $P-$value is the number (or interval) in Table B's column title that corresponds to the $t-$value in row localid="1650366573990" $n-1=11-1$

$=10$:

$0.05<P<0.10$

The null hypothesis is rejected if the P-value is less than the significance level.

$P>0.05=5\%\Rightarrow \text{Fail to reject}{H}_0$

There is not sufficient evidence to support the claim.