Chapter 26, Problem 12E

a)

To determine

To explain which test would be appropriate.

Answer to Problem 12E

Chi-square goodness of fit.

Explanation of Solution

Given:

  

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.

b)

To determine

To find the expected value for each cell.

Explanation of Solution

Given:

n = 655

p = proportion = $\frac{1}{49}$

Formula:

  $E=np$

Calculation:

The expected values are:

  $E=655\times \frac{1}{49}=13.3673$

c)

To determine

To state null and alternative hypotheses.

Answer to Problem 12E

  $\begin{array}{l}{H}_0:p_1=p_2=\mathrm{...}=p_{49}=\frac{1}{49}\\ H_a:Atleastoneofthe{p}_i’sdifferent\end{array}$

Explanation of Solution

Given:

  

The null and alternative hypotheses are:

  $\begin{array}{l}{H}_0:p_1=p_2=\mathrm{...}=p_{49}=\frac{1}{49}\\ H_a:Atleastoneofthe{p}_i’sdifferent\end{array}$

d)

To determine

To find the degrees of freedom.

Answer to Problem 12E

The degrees of freedom = 48

Explanation of Solution

Given:

  

The degrees of freedom are:

  $df=c-1=49-1=48$

e)

To determine

To find the p-value.

Answer to Problem 12E

The p-value is 0.9218

Explanation of Solution

Given:

  

  ${\chi }^2$ = 34.5

The degrees of freedom are 48

The p-value = 0.9218

Using excel formula,

=CHIDIST(34.5,48)

f)

To determine

To state the conclusion.

Answer to Problem 12E

There is not sufficient evidence to reject the claim that all numbers are equally likely.

Explanation of Solution

Given:

  ${\chi }^2$ = 34.5

P-value = 0.9218

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

Conclusion: There is not sufficient evidence to reject the claim that all numbers are equally likely.