Chapter 15, Problem 15.55SE

a.

To determine

To find:whether the sufficient evidence exist to indicate a difference in the mean response time for two stimuli.

Answer to Problem 15.55SE

No.

Explanation of Solution

Given:

An experiment was conducted to compare the response time for two stimuli.

To remove natural person to person variability in the response, both stimuli were presented to each of the nine subjects.

The given below table is response times (in second)

  

Calculation:

  $\begin{array}{l}n=\text{sample size}-\text{ties}=9-0=9\\ \alpha =\text{significant level}=0.05\end{array}$

The null hypothesis states that there is no difference in the populations. The alternatives hypothesis states the opposite of the null hypothesis.

  $\begin{array}{l}{H}_0:\text{ There is no difference between the two population distributions}\\ H_a:\text{ There is a difference between the two population distributions}\end{array}$

Determine the difference in the data in the data values of each pair and determine the sign of the difference. Difference of 0 are ignored.

First sample	Second sample	Difference	Sign
9.4	10.3	$9.4-10.3=0.9$	$-$
7.8	8.9	$7.8-8.9=-1.1$	$-$
5.6	4.1	$5.6-4.1=1.5$	$+$
12.1	14.7	$12.1-14.7=-2.6$	$-$
6.9	8.7	$6.9-8.7=-1.8$	$-$
4.2	7.1	$4.2-7.1=-2.9$	$-$
8.8	11.3	$8.8-11.3=-2.5$	$-$
7.7	5.2	$7.7-5.2=2.5$	$+$
6.4	7.8	$6.4-7.8=-1.4$	$-$

The observe value of the test statistics is then the number of positives signs.

  $x=2$

The $P\text{-value}$ is the probability of obtaining a value more extreme or equal the standardized test statistics. It is the fact the probability $P\left(x\le k\right)$ is given in the row $k$ and in the column $p=0.5$ . if the binomial cumulative table with $n$ in the appendix.

  $\begin{array}{l}P=P\left(x\le 2\text{or}x\ge n-2\right)=P\left(x\le 2\text{or}x\ge 7\right)\\ =2P\left(x\le 2\right)\\ =2\left(0.09\right)\\ =0.18\end{array}$

If the $P\text{-value}$ is less than the significance level $\alpha$ , then rejected the null hypothesis:

  $0.180>0.05\Rightarrow \text{Fail to reject }{H}_0$

There is not sufficient evidence to reject the claim that two schools are the same in academic effectiveness.

b.

To determine

To test: the hypothesis of no difference in the mean response time using student’s $t\text{-test}$ .

Answer to Problem 15.55SE

There is not sufficient evidence to support the claim that there is a difference in the mean response time for the two stimuli.

Explanation of Solution

Given:

An experiment was conducted to compare the response time for two stimuli.

To remove natural person to person variability in the response, both stimuli were presented to each of the nine subjects.

Calculation:

First to find the difference between each pair of data values

First sample	Second sample	Difference
9.4	10.3	$9.4-10.3=-0.9$
7.8	8.9	$7.8-8.9=-1.1$
5.6	4.1	$5.6-4.1=1.5$
12.1	14.7	$12.1-14.7=-2.6$
6.9	8.7	$6.9-8.7=-1.8$
4.2	7.1	$4.2-7.1=-2.9$
8.8	11.3	$8.8-11.3=-2.5$
7.7	5.2	$7.7-5.2=2.5$
6.4	7.8	$6.4-7.8=-1.4$

Now, find the sample mean of the differences. The mean is the sum of all values divided by the number of values.

  $\begin{array}{l}\overline{d}=\frac{-0.9-1.1-1.5-2.6-1.8-2.9-2.5+2.5-1.4}{9}\\ \overline{d}\approx -1.0222\end{array}$

The variance is the sum of squared deviations from the mean divided by $n-1$ . The standard deviation is the square root of the variance.

Therefore, the sample standard deviation of the differences:

  $\begin{array}{l}\overline{d}=\sqrt{\frac{\begin{array}{l}{\left(-0.9-\left(-1.0222\right)\right)}^2+{\left(-1.1-\left(-1.0222\right)\right)}^2+{\left(1.5-\left(-1.0222\right)\right)}^2+{\left(-2.6-\left(-1.0222\right)\right)}^2+\\ {\left(-1.8-\left(-1.0222\right)\right)}^2+{\left(-2.9-\left(-1.0222\right)\right)}^2+{\left(-2.5-\left(-1.0222\right)\right)}^2+\left(2.5-\left(-1.0222\right)\right)+{\left(-1.4-\left(-1.0222\right)\right)}^2\end{array}}{9-1}}\\ \overline{d}\approx 1.8620\end{array}$

the claim is either the null hypothesis or the alternative hypothesis. The null hypothesis needs to contain an equality. If the null hypothesis the claim, then the alternative hypothesis states that opposite of the null hypothesis.

  $\begin{array}{l}{H}_0:{\mu }_d=0\\ H_{\alpha }:{\mu }_d\ne 0\end{array}$

now, to find the value of the test statistic:

  $t=\frac{\overline{d}-{\mu }_d}{s_d/\sqrt{9}}=\frac{-1.022-0}{1.8620/\sqrt{9}}\approx -1.647$

The $P\text{-value}$ is the probability of obtaining a value of the test statistic, or a value more extreme, assuming the null hypothesis is true. The $P\text{-value}$ is the number (or interval) of column title of the students. T distribution in the appendix containing t-value in the row $df=n-1=9-1=8$ .

  $0.10=2\times 0.05<P<2\times 0.10=0.20$

If the $P\text{-value}$ is less than the significance level $\alpha$ , then rejected the null hypothesis:

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

There is not sufficient evidence to support the claim that there is a difference in the mean response time for the two stimuli.