Chapter 10.1, Problem 10.11E

(a)

To determine

To Explain: the appropriate hypotheses for a significance test to determine whether calculator’s random number generator is working properly.

Answer to Problem 10.11E

  $H_0:p_0=p_1=p_2=p_3=p_4=p_5=p_6=p_7=p_8=p_9=\frac{1}{10}=0.1$

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

Explanation of Solution

Given:

  $n=200$

Suppose:

  $\alpha =0.05$

If the digits from 0 to 9 are arbitrary generated, then each of the 10 digits require to have 1 chance in 10 to be choose.

  $p=\frac{1}{10}=0.1$

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

  $H_0:p_0=p_1=p_2=p_3=p_4=p_5=p_6=p_7=p_8=p_9=\frac{1}{10}=0.1$

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

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

(b)

To determine

To Calculate: a goodness of fit test and give expected counts, chi-square statistic, P-value, and conclusion.

Explanation of Solution

Formula used:

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

Calculation:

GENERATING SAMPLE

Arbitrary generating 200 random digits from 0 to 9 using the $randInt$ function

  $randInt\left(0,9,200\right)$

A possible result is then brief in the following table:

Digit	Observed Frequency   $O$
0	18
1	20
2	18
3	26
4	17
5	23
6	16
7	17
8	23
9	22

GOODNESS OF FIT TEST

The expected frequencies $E$ are the multiplication of the sample size and the probabilities

  $\begin{array}{c}{\text{E}}_1=n{p}_1=200\times 0.1=20\\ {\text{E}}_2=n{p}_2=200\times 0.1=20\\ E_3=n{p}_3=200\times 0.1=20\\ E_4=n{p}_4=200\times 0.1=20\\ E_5=n{p}_5=200\times 0.1=20\\ E_6=n{p}_6=200\times 0.1=20\\ E_7=n{p}_7=200\times 0.1=20\\ E_8=n{p}_8=200\times 0.1=20\\ E_9=n{p}_9=200\times 0.1=20\\ E_{10}=n{p}_{10}=200\times 0.1=20\end{array}$

The value of the test- statistic is

  $\begin{array}{c}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\\ =\frac{{\left(18-20\right)}^2}{20}+\frac{{\left(20-20\right)}^2}{20}\\ +\frac{{\left(18-20\right)}^2}{20}+\frac{{\left(26-20\right)}^2}{20}\\ +\frac{{\left(17-20\right)}^2}{20}+\frac{{\left(23-20\right)}^2}{20}\\ +\frac{{\left(16-20\right)}^2}{20}+\frac{{\left(17-20\right)}^2}{20}\\ +\frac{{\left(23-20\right)}^2}{20}+\frac{{\left(22-20\right)}^2}{20}\\ =5\end{array}$

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

  $df=c-1=10-1=9$

The P-Value is the probability of getting the value of the test statistic, or a value more extreme. The P-value is the number in the column of the chi-square distribution table containing the $x^2$ -value in the row $df=c-1=10-1=9$

  $p>0.25$

If the P-value is less than or equal to the significance level, then the 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.