Chapter 14.1, Problem 9E

a.

To determine

Find the five class intervals.

Answer to Problem 9E

The five class intervals are

$\left[0,0.2231\right),\left[0.2231,0.5108\right),\left[0.5108,0.9163\right),\left[0.9163,1.6094\right),\left[1.6094,\infty \right)$

Explanation of Solution

Given info:

The response time of a computer system follows exponential distribution with parameter 1.

The probability density function is given below:

$f_0\left(x\right)=e^{-x}\text{for }x\ge 0$

Calculation:

The total area covered equals one. The area covered by each of the five intervals would be 0.2.

Denote the five class intervals by $c_1,c_2,c_3,c_4,c_5$.

As x is greater than or equal to zero.

The class limits for the first interval would be $\left[0,{\text{c}}_{\text{1}}\right)$

$\begin{array}{c}P\left(0\le X\le c_1\right)={\displaystyle \underset{0}{\overset{c_1}{\int }}{e}^{-x}}dx\\ ={\left(-e^{-x}\right)}_0^{c_1}\\ =1-e^{-c_1}\\ 0.2=1-e^{-c_1}\end{array}$

$\begin{array}{c}{c}_1=-\ln \left(0.8\right)\\ =0.2231\end{array}$

The class limits for the second class would be $\left[c_1,c_2\right)$. The area covered by the second interval would be 0.2. Also, area below c₂ is 0.4

$\begin{array}{c}P\left(c_1\le X\le c_2\right)={\displaystyle \underset{c_1}{\overset{c_2}{\int }}{e}^{-x}}dx\\ ={\displaystyle \underset{0}{\overset{c_2}{\int }}{e}^{-x}}dx\\ ={\left(-e^{-x}\right)}_0^{c_2}\\ =1-e^{-c_2}\\ 0.4=1-e^{-c_2}\end{array}$

$\begin{array}{c}{c}_2=-\ln \left(0.6\right)\\ =0.5108\end{array}$

The class limits for the third class would be $\left[c_2,c_3\right)$. The area covered by the third interval would be 0.2. Also, area below c₃ is 0.6

$\begin{array}{c}P\left(c_2\le X\le c_3\right)={\displaystyle \underset{c_2}{\overset{c_3}{\int }}{e}^{-x}}dx\\ ={\displaystyle \underset{0}{\overset{c_3}{\int }}{e}^{-x}}dx\\ ={\left(-e^{-x}\right)}_0^{c_3}\\ =1-e^{-c_3}\\ 0.6=1-e^{-c_3}\end{array}$

$\begin{array}{c}{c}_3=-\ln \left(0.4\right)\\ =0.9163\end{array}$

The class limits for the fourth class would be $\left[c_3,c_4\right)$. The area covered by the fourth interval would be 0.2. Also, area below c₄ is 0.4

$\begin{array}{c}P\left(c_3\le X\le c_4\right)={\displaystyle \underset{c_3}{\overset{c_4}{\int }}{e}^{-x}}dx\\ ={\displaystyle \underset{0}{\overset{c_4}{\int }}{e}^{-x}}dx\\ ={\left(-e^{-x}\right)}_0^{c_4}\\ =1-e^{-c_4}\\ 0.8=1-e^{-c_4}\end{array}$

$\begin{array}{c}{c}_4=-\ln \left(0.2\right)\\ =1.6094\end{array}$

Thus, the five class intervals are $\left[0,0.2231\right),\left[0.2231,0.5108\right),\left[0.5108,0.9163\right),\left[0.9163,1.6094\right),\left[1.6094,\infty \right)$

b.

To determine

Carry out chi-square test for the given data.

Answer to Problem 9E

There is sufficient evidence to conclude that the data is consistent with the exponential distribution.

Explanation of Solution

Given info:

The data shows the 40 response times.

Calculation:

The observed frequency for each class interval is given below:

Class interval	Frequency
0-0.2231	6
0.2231-0.5108	8
0.5108-0.9163	10
0.9163-1.6094	7
1.6094 or more	9

Testing the hypothesis:

Null hypothesis:

$H_0:p_1=p_2=p_3=p_4=p_5=0.2$

Alternative hypothesis:

$H_a:$ At least one of the observed proportions is not equal to the expected proportion.

Expected frequency:

The expected frequency for each group is calculated as follows,

$\text{Expected frequency}=n{p}_i$

Where,

n is the total number of observed frequency.

$p_i$ is the proportion corresponding to a particular group.

As the expected proportion for all the five class intervals are same, the expected frequency would be a single value and it is calculated as follows:

$\begin{array}{c}\text{Expected frequency}=\left(40\right)\left(0.2\right)\\ =0.8\end{array}$

Test statistic:

${\chi }^2={\displaystyle \sum _i\frac{{\left(n_i-n{p}_{i0}\right)}^2}{n{p}_{i0}}}$

Where,

$n_i$ represents the observed frequency.

$n{p}_i$ represents the expected frequency.

The table shows the calculation for chi-square test statistic:

Class interval	Observed $n_i$	Expected $n{p}_i$	${\chi }^2={\displaystyle \sum _i\frac{{\left(n_i-n{p}_{i0}\right)}^2}{n{p}_{i0}}}$
0-0.2231	6	8	0.5
0.2231-0.5108	8	8	0
0.5108-0.9163	10	8	0.5
0.9163-1.6094	7	8	0.125
1.6094 or more	9	8	0.125
Total			1.25

Thus, the test statistic is 1.25.

Degrees of freedom:

If there are k categories given, then the degrees of freedom would be $k-1$

Here, there are five categories, the degrees of freedom is,

$\begin{array}{c}k-1=5-1\\ =4\end{array}$

Critical value:

Use the Table A.7 to find the chi-square critical value.

• Locate 4 under the column of v.
• In the row of $\alpha$ locate the value0.10.
• The value intersecting these two numbers gives the critical value corresponding to ${\chi }_{0.10,4}^2$

Thus, the critical value is ${\chi }_{0.10,4}^2$ is7.779.

The level of significance is taken as 0.10 because the critical value greater than 0.10 would result in 1.064 or less and critical value for0.05would result 9.488 or more. Hence, the level of significance was taken as 0.10.

Decision rule:

The null hypothesis would be rejected if the P-value is lesser than or equal to the level of significance $\alpha$ and this would occur if the test statistic of chi-square is greater than or equal to the critical value.

Conclusion:

The test statistic value is 1.25 and the critical value is 7.779.

The test statistic value is lesser than the critical value.

Hence, the P-value would be greater than the level of significance.

That is, $P\text{-value}>0.10\left(=\alpha \right)$

Hence, the null hypothesis isnot rejected.

Thus, there is sufficient evidence to conclude that the data is consistent with the exponential distribution.