Chapter 9, Problem 2RE

a)

To determine

To construct a boxplot

Explanation of Solution

Given:

$x$
2	5
10	9
3	15
9	15
15	4
10	5
7	13
4	6
3	14
7	11
5	6
9	12

Boxplot for this data is,

  

Above boxplot shows median is approximately middle of the box as well as tails of the box are at same distance from the box. So, this boxplot shows symmetry of the data. Hence it can be assumed that, this population follows approximately normal distribution.

Assumption of testing of hypothesis are satisfied

b)

To determine

To perform hypothesis testing

Explanation of Solution

First need to find mean and standard deviation from the data.

Formula:

Mean:

  $\overline{x}=\frac{\sum x_i}{n}$

Standard deviation:

  $s=\sqrt{\frac{{\displaystyle \sum {(x_i-\overline{x})}^2}}{n-1}}$

Test statistic:

  $t=\frac{\overline{x}-\mu }{s/\sqrt{n}}$

Calculation:

Creating table to find mean and sample standard deviation:

X
2	-6.29	39.56
10	1.71	2.92
3	-5.29	27.98
9	0.71	0.50
15	6.71	45.02
10	1.71	2.92
7	-1.29	1.66
4	-4.29	18.40
3	-5.29	27.98
7	-1.29	1.66
5	-3.29	10.82
9	0.71	0.50
5	-3.29	10.82
9	0.71	0.50
15	6.71	45.02
15	6.71	45.02
4	-4.29	18.40
5	-3.29	10.82
13	4.71	22.18
6	-2.29	5.24
14	5.71	32.60
11	2.71	7.34
6	-2.29	5.24
12	3.71	13.76

Mean:

  $\overline{x}=\frac{199}{24}$

  $\overline{x}$ = 8.29

Standard deviation:

  $s=\sqrt{\frac{396.96}{23}}=4.15$

From the data, sample mean is $\overline{x}$ = 8.29 and the Sample standard deviation is s = 4.15, and the sample size is n = 24.

Null and Alternative Hypotheses:

  $\begin{array}{l}{H}_0:\mu =8.62\\ H_a:\mu <8.62\end{array}$

This corresponds to a left-tail test, for which a t-test for one mean, with unknown population standard deviation will be used.

  $t=\frac{8.29-8.62}{4.15/\sqrt{24}}$

t = -0.39

The number of degrees of freedom are df = n-1 = 24-1=23

Rejection Region:

Given significance level = a = 0.05 and df = 23

So Critical Value for the test is, $t_c=$ -1.714 ...Using t-table

Decision:

Since it is observed that |t| = 0.39< $\left|t_c\right|=$ 1.714

It is then concluded that the Null Hypothesis is not rejected.

Conclusion: It is concluded that the Null Hypothesis is not rejected. Therefore, there is not enough evidence to claim that the Mean number of runs in 2013 is less than it was in 2012 at the 0.05 significance level