MHB Is the median speed on this highway different from 55 mph?

[MarkFL]

Hello MHB Community,

anemone is a bit under the weather this week, so she has asked me to fill in for her. Please join me in wishing for her a speedy recovery. (Yes)

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A highway official wants to determine whether the average speed by drivers on a certain section of highway differs from the posted limit of 55 mph. Twelve randomly selected cars are monitored for speed. The speeds, in miles per hour, are:

$$\left[\begin{array}{c}\hline65 & 58 & 54 & 50 \\ 48 & 60 & 54 & 64 \\ 57 & 61 & 55 & 56 \\\hline \end{array}\right]$$

Can we conclude that the median speed by drivers on this section of highway is different from 55 mph? Employ the Wilcoxin signed-rank test with a significance level as close to 0.05 as possible. [Assume that the speeds have a symmetric distribution].

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Hi MHB,

I would like to genuinely thank MarkFL for stepping up and helping me in taking care of the last week POTW duties when I was sick, your help is much appreciated!(Yes):)

Congratulations to laura123 for the correct solution:

Spoiler

Let'us employ the Wilcoxon signed-rank test with a significance level $\alpha=0.05$.
Let $m$ be the median speed:
$H_0: m=55\mbox{ mph}$ versus $H_1: m\neq 55\mbox{ mph}$
\begin{array}{ |c|c|c|c|c| } \hline \hline x_i & x_i-m & |x_i-m| & rank & R_+ \\ \hline\hline 65&10 &10 & 11&11 \\\hline 58&3 &3 & 5&5 \\\hline 54&-1 &1 & 2&- \\\hline 50&-5 &5 & 6.5&- \\\hline 48&-7 &7 & 9&- \\\hline 60&5 &5 & 6.5&6.5 \\\hline 54&-1 &1 & 2&- \\\hline 64&9 &9 & 10&10 \\\hline 57&2 &2 & 4&4 \\\hline 61&6 &6 & 8&8 \\\hline 55&0 &0 & -&- \\\hline 56&1 &1 & 2&2 \\\hline\end{array}
$W=\sum R_+=11+5+6.5+10+4+8+2=46.5$
This is a two tail test. For $n=11$ and $\alpha=0.05$ the lower and upper critical values for $W$ are $W_L=11$ and $W_U=55$.
$W_L<46.5<W_U$ so $H_0$ cannot be rejected al the 0.05 level.

Here is the suggested solution provided by MarkFL:

Spoiler

Step 1: State the null and alternative hypotheses.

Let $\eta$ denote the median speed by drivers on the section of highway. Then the null and alternative hypotheses are:

[box=gray]$$H_0:\,\eta=\text{55 mph (median speed is 55 mph)}$$
$$H_a:\,\eta\ne\text{55 mph (median speed is not 55 mph)}$$[/box]

Note that the hypothesis test is two-tailed since there is a not-equal sign ($\ne$) in the alternate hypothesis.

Step 2: Decide on a significance level and use a table showing the critical values and significance levels for a Wilcoxin signed-rank test to find a significance level, $\alpha$, as close as possible to the one required.

To begin, note that there is a data value in the sample that equals the value , 55, given for the median in the null hypothesis; nemaly, the third entry in the third row of the given data table. This data value must be deleted from the sample. This reduces the sample size from 12 to 11. Now we look in the critical values table for a significance level as close as possible to 0.05 for a two-tailed test with $n=11$. From the table, we see that the best we can do is [m]$\alpha=0.054$[/m].

Step 3: The critical values for a two-tailed test are $W_\ell$ and $W_r$.

Consulting the critical values table, we find that the critical values for a two-tailed test with $n=11$ and $\alpha=0.054$ are [m]$W_\ell=11$ and $W_r=55$[/m].

Step 4: Construct a work table.

Referring to the given data table and recalling that the data value 55 has been deleted, we obtain the following work table:

[TABLE="class: grid, width: 600, align: center"]
[TR]
[TD]Speed (mph)
x[/TD]
[TD]Difference
$D=x-55$[/TD]
[TD]$|D|$[/TD]
[TD]Rank of $|D|$[/TD]
[TD]Signed rank R[/TD]
[/TR]
[TR]
[TD]65[/TD]
[TD]10[/TD]
[TD]10[/TD]
[TD]11[/TD]
[TD]11[/TD]
[/TR]
[TR]
[TD]48[/TD]
[TD]-7[/TD]
[TD]7[/TD]
[TD]9[/TD]
[TD]-9[/TD]
[/TR]
[TR]
[TD]57[/TD]
[TD]2[/TD]
[TD]2[/TD]
[TD]4[/TD]
[TD]4[/TD]
[/TR]
[TR]
[TD]58[/TD]
[TD]3[/TD]
[TD]3[/TD]
[TD]5[/TD]
[TD]5[/TD]
[/TR]
[TR]
[TD]60[/TD]
[TD]5[/TD]
[TD]5[/TD]
[TD]6.5[/TD]
[TD]6.5[/TD]
[/TR]
[TR]
[TD]61[/TD]
[TD]6[/TD]
[TD]6[/TD]
[TD]8[/TD]
[TD]8[/TD]
[/TR]
[TR]
[TD]54[/TD]
[TD]-1[/TD]
[TD]1[/TD]
[TD]2[/TD]
[TD]-2[/TD]
[/TR]
[TR]
[TD]54[/TD]
[TD]-1[/TD]
[TD]1[/TD]
[TD]2[/TD]
[TD]-2[/TD]
[/TR]
[TR]
[TD]50[/TD]
[TD]-5[/TD]
[TD]5[/TD]
[TD]6.5[/TD]
[TD]-6.5[/TD]
[/TR]
[TR]
[TD]64[/TD]
[TD]9[/TD]
[TD]9[/TD]
[TD]10[/TD]
[TD]10[/TD]
[/TR]
[TR]
[TD]56[/TD]
[TD]1[/TD]
[TD]1[/TD]
[TD]2[/TD]
[TD]2[/TD]
[/TR]
[/TABLE]

Note that the 1's in the $|D|$-column are all tied for smallest. Hence, each is assigned the rank $$\frac{1+2+3}{3}=2$$. Also, the two 5's in the $|D|$-column are tied for sixth smallest, so each is assigned the rank $$\frac{6+7}{2}=6.5$$.

Step 5: Compute the value of the test statistic.

$$W=\text{sum of the positive ranks}$$​

From the final column of our work table, we see that the sum of the positive ranks is equal to:

$$W=11+4+5+6.5+8+10+2=46.5$$​

Step 6: If the value of the test statistic falls in the rejection region, reject $H_0$; otherwise, do not reject $H_0$.

The value of the test statistic, $W=46.5$, does not fall in the rejection region . Thus, we do not reject $H_0$.

Step 7: State the conclusion.

The data do not provide sufficient evidence to conclude that the median speed differs from 55 mph.