Chapter 11.1, Problem 18E

(a)

To determine

To Explain: the appropriate hypotheses for a chi-square test for goodness of fit to fine whether the random number generator give every digit an equal possibility of being generated.

Answer to Problem 18E

  $H_0:p_0=p_1=p_2=p_3=p_4=p_5=p_6=p_7=p_8=p_9=0.1$

  $H_1:{\text{ At least one of the P}}_i\text{'}\text{s is incorrect}$

Explanation of Solution

Every digit has equal possible results and there are 10 digits. Therefore the probability of arbitrary digit is 1 out of 10:

  $p=\frac{1}{10}=0.1$

The null hypothesis states that the specified distribution of the categorical variable is correct.

  $H_0:p_0=p_1=p_2=p_3=p_4=p_5=p_6=p_7=p_8=p_9=0.1$

The alternative hypothesis statement is that the specified distribution of the categorical variable is not correct.

  $H_1:{\text{ At least one of the P}}_i\text{'}\text{s is incorrect}$

(b)

To determine

To find: out the test at the significance level.

Answer to Problem 18E

There is not enough proof to reject the claim of random digits.

Explanation of Solution

Find the observed frequencies and the chi-square subtotals.

O represents the observed frequencies found from entering randInt(0,9, 200)

Into the calculator

Distribution	O	E = np	O-E	(O-E)2	(O-E)2/E
0.1	22	20	2	4	0.2
0.1	21	20	1	1	0.05
0.1	24	20	4	16	0.8
0.1	15	20	-5	25	1.25
0.1	17	20	-3	9	0.45
0.1	20	20	0	0	0
0.1	19	20	-1	1	0.05
0.1	21	20	1	1	0.05
0.1	18	20	-2	4	0.2
0.1	23	20	3	9	0.45

The test-statistic is

  ${\chi }^2=3.5$

The degree of freedom is

  $df=c-1=10-1=9$

The P-value is the probability of getting the value of the test statistic, or a value more than extreme. The P-value is the number in the column of Table having the ${\chi }^2$ -value in the row $df=9$ :

  $P>0.25$

If the P-value is equal or lesser the significance level, then the null hypothesis is rejected:

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

(c)

To determine

To find: the probability that the student would make a Type I error in part (b).

Answer to Problem 18E

0.05

Explanation of Solution

The probability of a Type I error is the $\alpha$ -value. Therefore, the possibility of a Type I error is $\alpha =0.05$

(d)

To determine

To find: the probability that at least one of them makes a Type I error.

Explanation of Solution

Formula used:

Multiplication rule

  $P\left(A\cap B\right)=P\left(A\text{ and B}\right)=P\left(A\right)\times P\left(B\right)$

Complement rule:

  $P\left(A^c\right)=P\left(\text{ not }A\right)=1-P\left(A\right)$

Calculation:

Result part (c):

P (Type I error) = 0.05

Using the complement rule:

P (No Type I error) = 1 − P (Type I error) = 1- 0.05 = 0.95

Suppose that the 25 students are independent, can use the multiplication rule for independent events:

  $\begin{array}{c}P\left(\text{None of the 25 make type I error}\right)=\underset{25\text{ repetitions}}{\underbrace{P\left(\text{No Type I error}\right)\times P\left(\text{No Type I error}\right)\times \mathrm{...}\times P\left(\text{No Type I error}\right)}}\\ ={\left(P\left(\text{No Type I error}\right)\right)}^{25}\\ ={0.95}^{25}\\ =0.2774\end{array}$

Using the compliment rule

  $\begin{array}{c}P\left(\text{At least one of the 25 make type I error}\right)=1-P\left(\text{None of the 25 make type I erro}\right)\\ =1-0.2774\\ =0.7226\end{array}$