Chapter 10.1, Problem 10.12E

To determine

To Explain: a significance test of the hypothesis that this program yields an equal proportion of 1s, 2s, 3s and 4s.

Answer to Problem 10.12E

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

There is not enough evidence to reject the claim that the digits were randomly selected.

Explanation of Solution

Given:

  $n=200$

Formula used:

  $x^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}$

Calculation:

Assume:

  $\alpha =0.05$

One of the four regions is arbitrary chose, and then every of the 4 regions requires to have 1 chance in four to be selected.

  $p=\frac{1}{4}=0.25$

The null hypothesis statement is the population proportions are equal to the given proportions:

  $H_0:p_0=p_1=p_2=p_3=p_4=\frac{1}{4}=0.25$

The alternative hypothesis statement is the opposite of the null hypothesis:

  $H_1:{\text{At least one of the p}}_i\text{'s is different}\text{.}$

Randomly generating the sample

A possible result is then in brief

Region	Observed Frequency   $O$
1	49
2	50
3	47
4	54

GOODNESS OF FIT TEST

The multiplication of the sample size and the probabilities is the expected frequencies are

  $\begin{array}{c}{\text{E}}_1=n{p}_1=200\times 0.25=50\\ {\text{E}}_2=n{p}_2=200\times 0.25=50\\ E_3=n{p}_3=200\times 0.25=50\\ E_4=n{p}_4=200\times 0.25=50\end{array}$

The value of the test- statistic is

  $\begin{array}{c}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\\ =\frac{{\left(49-50\right)}^2}{50}+\frac{{\left(50-50\right)}^2}{50}\\ +\frac{{\left(47-50\right)}^2}{50}+\frac{{\left(54-50\right)}^2}{50}\\ =0.52\end{array}$

The degree of freedom is the number of variables decreased by 1.

  $df=c-1=4-1=3$

The P-Value is the probability of obtaining the value of the test statistic, or value more extreme. The P-value is the number in the column of the chi-square distribution containing the $x^2$ -value in the row $df=c-1=4-1=3:$

  $P>0.25$

If the P-value is equal or less than to the significance level, then the Alternate hypothesis is accepted and null hypothesis is rejected:

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

There is no enough evidence to reject the claim that the digits were randomly selected.