Chapter 10, Problem R10.7RE

(a)

To determine

To construct a dot plot of the difference (Standing minus blocks) in $50$ -meter run time for each sprinter and explain what does the graph suggest about whether starting blocks are helpful.

Explanation of Solution

Firstly, let us find the difference between the time with blocks and time in standing start for each sprinter, i.e.

Sprinter	With Blocks	Standing Blocks	Difference
1	6.12	6.38	-0.26
2	6.42	6.52	-0.1
3	5.98	6.09	-0.11
4	6.8	6.72	0.08
5	5.73	5.98	-0.25
6	6.04	6.27	-0.23
7	6.55	6.71	-0.16
8	6.78	6.8	-0.02

Now we will create a dot plot as:

We note that 7 out of 8 dots lie to the left of zero, which indicates that most of the differences are negative and thus that most of the times blocks are less than the times in standing start.

This then implies that the graph suggests that starting blocks are helpful and reduce time. Thus, the graph suggest that the starting blocks are helpful.

(b)

To determine

To calculate the mean difference and the standard deviation of the differences and explain why mean differences gives some evidence that the starting blocks are equal.

Answer to Problem R10.7RE

The mean is $-0.1312$ .

The standard deviation is $0.1193$ .

Explanation of Solution

As in part (a), we have find the difference between the time and blocks in starting start for each sprinter, we have,

Sprinter	With Blocks	Standing Blocks	Difference
1	6.12	6.38	-0.26
2	6.42	6.52	-0.1
3	5.98	6.09	-0.11
4	6.8	6.72	0.08
5	5.73	5.98	-0.25
6	6.04	6.27	-0.23
7	6.55	6.71	-0.16
8	6.78	6.8	-0.02

The mean of the difference will be as:

  $\begin{array}{l}{\overline{x}}_d=\frac{-0.26-0.1-0.11+0.08-0.25-0.23-0.16-0.02}{8}\\ =\frac{-1.05}{8}\\ =-0.1312\end{array}$

Thus the mean is $-0.1312$ .

The standard deviation is the square root of the variance then,

  $\begin{array}{l}{s}_d=\sqrt{\frac{\begin{array}{}$ {(-0.26+0.1312)}^2+{(-0.1+0.1312)}^2+{(-0.11+0.1312)}^2+{(0.08+0.1312)}^2+{(-0.25+0.1312)}^2$$ +{(-0.23+0.1312)}^2+{(-0.16+0.1312)}^2+{(-0.02+0.1312)}^2$\end{array}}{8-1}}\\ =0.1193\end{array}$

The standard deviation is $0.1193$ .

Since the sample mean of the difference $-0.1312$ is negative this indicates that the mean time with blocks is lower than the mean time with standing start and thus this indicates that the starting blocks are helpful, which implies that the mean differences give some evidence that starting blocks are helpful.

(c)

To determine

To find out do the data provide the convincing evidence that sprinters like these runs a faster race when using starting blocks on average or not.

Answer to Problem R10.7RE

There is convincing evidence that sprinters like these run a faster race when using starting blocks, on average.

Explanation of Solution

It is given in the question that:

  $\begin{array}{l}n=8\\ \alpha =0.05\end{array}$

Now, from part (b), we know that,

The mean is $-0.1312$ and the standard deviation is $0.1193$ .

Thus, the hypothesis test will be as follows:

Claim given: Mean is lower for with blocks.

The claim is either the null hypothesis or the alternative hypothesis.

  $\begin{array}{l}{H}_0:{\mu }_d=0\\ H_a:{\mu }_d<0\end{array}$

Let us calculate the test statistics:

  $\begin{array}{l}t=\frac{\overline{d}-{\mu }_d}{\frac{s_d}{\sqrt{n}}}\\ =\frac{-0.1312-0}{\frac{0.1193}{\sqrt{8}}}\\ =-3.111\end{array}$

The degree of freedom will be:

  $\begin{array}{l}df=n-1\\ =8-1\\ =7\end{array}$

The P-values will be:

  $0.005<P<0.01$

If the P-value is less than the significance level, reject the null hypothesis:

Thus, we have, $P<0.05$ , which implies the null hypothesis is rejected.

So, we can conclude that there is convincing evidence that sprinters like these run a faster race when using starting blocks, on average.

(d)

To determine

To construct and interpret $90\%$ confidence interval for the true mean difference and explain how the confidence interval gives more information that the test in part (b).

Answer to Problem R10.7RE

The confidence interval is $(-0.2111,-0.0513)$ .

We are $90\%$ confident that the true mean difference is between $-0.2111\text{ seconds and -0}\text{.0513 seconds}$ .

Explanation of Solution

It is given in the question that:

  $\begin{array}{l}n=8\\ c=0.90\end{array}$

Now, from part (b), we know that,

The mean is $-0.1312$ and the standard deviation is $0.1193$ .

The degree of freedom will be:

  $\begin{array}{l}df=n-1\\ =8-1\\ =7\end{array}$

So, the t -value will be:

  $t_{\alpha /2}=1.895$

So, the margin of error will be:

  $\begin{array}{l}E=t_{\alpha /2}\times \frac{s}{\sqrt{n}}\\ =1.895\times \frac{0.1193}{\sqrt{8}}\\ =0.0799\end{array}$

Then the confidence interval will be calculated as:

  $\begin{array}{l}\overline{x}-E=-0.1312-0.0799\\ =-0.2111\\ \overline{x}+E=-0.1312+0.0799\\ =-0.0513\end{array}$

Thus, we are $90\%$ confident that the true mean difference is between $-0.2111\text{ seconds and -0}\text{.0513 seconds}$ .

The confidence interval gives more information than the hypothesis test because the confidence interval gives a range of possible values for the mean difference while the hypothesis test only tests a claim about one single value for the mean difference.