## 17.45

Right versus left. The design of controls and instruments affects how easily people can use them.  Timothy Sturm investigated this effect in a course project, asking 25 right-handed students to turn a knob (with their right hands) that moved an indicator by screw action.  There were two identical instruments, one with a right-hand thread (the knob turns clockwise) and the other with a left-hand thread (the knob turns counterclockwise).  The table below gives the times in seconds each subject took to move the indicator a fixed distance.

$\begin{array}{|c|rrrrrrrrrr|}\hline\text{subject}&1&2&3&4&5&6&7&8&9&10\\ \hline\text{right thread}&113&105&130&101&138&118&87&116&75&96\\ \text{left thread}&137&105&133&108&115&170&103&145&78&107\\ \hline\hline\text{subject}&11&12&13&14&15&16&17&18&19&20\\ \hline\text{right thread}&122&103&116&107&118&103&111&104&111&89\\ \text{left thread}&84&148&147&87&166&146&123&135&112&93\\ \hline\hline\text{subject}&21&22&23&24&25&&&&&\\ \hline\text{right thread}&78&100&89&85&88&&&&&\\ \text{left thread}&76&116&78&101&123&&&&&\\ \hline\end{array}$

## Solution

Since we want to prove that one of the times is less than the other, it makes sense to rewrite the previous table as one where we only indicate, for each subject, the difference between the time to turn the left thread, minus the time to turn the right thread.  We want to prove that the mean of this time is actually positive:

$\begin{array}{|c|rrrrrrrrrr|}\hline\text{subject}&1&2&3&4&5&6&7&8&9&10\\ \hline \text{difference}&24&0&3&7&-23&52&16&29&3&11\\ \hline \hline\text{subject}&11&12&13&14&15&16&17&18&19&20\\ \hline\text{difference}& -38&45&31&-20&48&43&12&31&1&4\\ \hline \hline\text{subject}&21&22&23&24&25&&&&&\\ \hline\text{difference}&-2&16&-11&16&35&&&&& \\ \hline \end{array}$

A carefully-split stemplot of the data shows a distribution close to Normal, although a bit skewed.  This is good news—intuitively—in our favor, since it seems to support our claims.  A histogram is also provided, for further visual verification:

 $\begin{array}{r|llllll} -3&8\\-2&0&3\\ -1&1\\ -0&2\\ 0&0&1&3&3&4&7\\ 1&1&2&6&6&6\\ 2&4&9\\ 3&1&1&5\\ 4&3&5&8\\ 5&2 \end{array}$

We assume that this is a proper SRS, so the use of t statistics is justified.  We want to prove that it takes longer to turn the left thread than the right, so to define our hypotheses, we set $\mu$ to be the mean difference, to be contrasted against $\mu_0=0$ (which indicates that left and right threads take the same time).  It should be then

$\begin{cases} H_0 &: \mu = 0 ,\\ H_a &: \mu >0. \end{cases}$

Notice that this is one-sided, as we make emphasis in one of the hands being faster than the other.  For the rest of the analysis, we will need both the mean and standard deviation of the SRS.  These values are respectively $\bar{x}= 13.32$ and $s=22.94$, as computed from the following table:

$\begin{array} {|r|r|r|r|} \hline k&x_k&x_k-\bar{x}&(x_k-\bar{x})^2\\ \hline \hline 1&24&10.68&114.0620\\ \hline 2&0&-13.32&177.4220\\ \hline 3&3&-10.32&106.5020\\ \hline 4&7&-6.32&39.9424\\ \hline 5&-23&-36.32&1345.4200\\ \hline 6&52&38.68&1496.1400\\ \hline 7&16&2.68&7.1824\\ \hline 8&29&15.68&245.8620\\ \hline 9&3&-10.32&106.5020\\ \hline 10&11&-2.32&5.3824\\ \hline 11&-38&-51.32&2633.7400\\ \hline 12&45&31.68&1003.6200\\ \hline 13&31&17.68&312.5820\\ \hline 14&-20&-33.32&1110.2200\\ \hline 15&48&34.68&1202.7000\\ \hline 16&43&29.68&880.9020\\ \hline 17&12&-1.32&1.7424\\ \hline 18&31&17.68&312.582\\ \hline 19&1&-12.32&151.7820\\ \hline 20&4&-9.32&86.8624\\ \hline 21&-2&-15.32&234.7020\\ \hline 22&16&2.68&7.1824\\ \hline 23&-11&-24.32&591.4620\\ \hline 24&16&2.68&7.1824\\ \hline 25&35&21.68&470.0220 \\ \hline \end{array}$

The one-sample t statistic is then

$t = \displaystyle{ \frac{\bar{x}-\mu_0}{s/\sqrt{n}} = \frac{13.32}{(22.94/5)} = 2.90322581}.$

In order to find the corresponding P-value, we have to compute the area to the right of 2.90 under the t–distribution curve with 24 degrees of freedom.  If we are to use Table C for this purpose, we need to pin P between the two closer values in that table.  We search the twenty-fourth row for those two entries that bracket 2.90 to get respectively 2.797 (99%, which gives us $P=1-0.99=0.01$) and 3.091 (99.5%, which gives $P=1-0.995=0.005$).  Since the table gives us the probabilities for the two-sided case, and we need the one-sided case, we need to divide those P-values by two.  The actual P–value is then between $0.0025$ and $0.005.$

The only detail left to finish the analysis is the decision of whether this P–value is small.  We have not given any significance threshold $\alpha$, but if we chose the usual thresholds $\alpha=0.05$ or $\alpha=0.1$, both our computed P–values are still much smaller than those.  We can thus guarantee that we have enough evidence to support the claim that right threads take less time than left threads.

1. October 27, 2010 at 11:11 pm

On a stem plot, where can I list 0?

• October 28, 2010 at 1:32 am

In the stem of zero (think 0 as 00). In the original stemplot that I posted, I forgot to include two pieces of data (the zero was one of those). Thanks to your comment I realized the typo, and corrected it.

2. March 25, 2011 at 8:37 pm

This was a GREATTTTTTT help!!! Thanks!!