Chapter 26, Problem 11E

a)

To determine

To find the expected value for each cell.

Explanation of Solution

Given:

n = 96

p = proportion = $\frac{1}{16}$

Formula:

  $E=np$

Calculation:

The expected values are:

  $E=96\times \frac{1}{16}=6$

Therefore, the table of observed and expected values becomes,

Observed	Expected
6	6
1	6
7	6
5	6
8	6
4	6
7	6
5	6
8	6
10	6
9	6
6	6
4	6
4	6
5	6
7	6
96	96

b)

To determine

To explain which test would be appropriate.

Answer to Problem 11E

Chi-square goodness of fit.

Explanation of Solution

Given:

Observed	Expected
6	6
1	6
7	6
5	6
8	6
4	6
7	6
5	6
8	6
10	6
9	6
6	6
4	6
4	6
5	6
7	6
96	96

The researcher wants to test whether the distribution is uniform. The test involves only one variable so, Chi-square goodness of fit would be appropriate to use.

c)

To determine

To state null and alternative hypotheses.

Answer to Problem 11E

  $\begin{array}{l}{H}_0:p_1=p_2=\mathrm{...}=p_{16}=\frac{1}{16}=0.0625\\ H_a:Atleastoneofthe{p}_i’sdifferent\end{array}$

Explanation of Solution

Given:

Observed	Expected
6	6
1	6
7	6
5	6
8	6
4	6
7	6
5	6
8	6
10	6
9	6
6	6
4	6
4	6
5	6
7	6
96	96

The null and alternative hypotheses are:

  $\begin{array}{l}{H}_0:p_1=p_2=\mathrm{...}=p_{16}=\frac{1}{16}=0.0625\\ H_a:Atleastoneofthe{p}_i’sdifferent\end{array}$

d)

To determine

To find the degrees of freedom.

Answer to Problem 11E

The degrees of freedom = 15

Explanation of Solution

Given:

Observed	Expected
6	6
1	6
7	6
5	6
8	6
4	6
7	6
5	6
8	6
10	6
9	6
6	6
4	6
4	6
5	6
7	6
96	96

The degrees of freedom are:

  $df=c-1=16-1=15$

e)

To determine

To find the ${\chi }^2$ and P-value.

Answer to Problem 11E

  ${\chi }^2$ = 12.67 and P-value = 0.6280

Explanation of Solution

Given:

Observed	Expected
6	6
1	6
7	6
5	6
8	6
4	6
7	6
5	6
8	6
10	6
9	6
6	6
4	6
4	6
5	6
7	6
96	96

Using excel,

observed	expected	O - E	(O - E)² / E	% of chisq
6	6.000	0.000	0.000	0.00
1	6.000	-5.000	4.167	32.89
7	6.000	1.000	0.167	1.32
5	6.000	-1.000	0.167	1.32
8	6.000	2.000	0.667	5.26
4	6.000	-2.000	0.667	5.26
7	6.000	1.000	0.167	1.32
5	6.000	-1.000	0.167	1.32
8	6.000	2.000	0.667	5.26
10	6.000	4.000	2.667	21.05
9	6.000	3.000	1.500	11.84
6	6.000	0.000	0.000	0.00
4	6.000	-2.000	0.667	5.26
4	6.000	-2.000	0.667	5.26
5	6.000	-1.000	0.167	1.32
7	6.000	1.000	0.167	1.32
Total = 96	96.000	0.000	12.667	100.00
12.67	chi-square
15	df
.6280	p-value

So,

  ${\chi }^2$ = 12.67

P-value = 0.6280

f)

To determine

To state the conclusion.

Answer to Problem 11E

There is not sufficient evidence to reject the claim that the frequency of such hurricanes has remained constant.

Explanation of Solution

Given:

  ${\chi }^2$ = 12.67

P-value = 0.6280

Decision: The p-value> 0.05, fail to reject H0.

Conclusion: There is not sufficient evidence to reject the claim that the frequency of such hurricanes has remained constant.