Q.14 Use your calculator鈥檚 RandInt ... [FREE SOLUTION]

91影视

The Practice of Statistics for AP

David Moore,Daren Starnes,Dan Yates

$Math Studyset 91影视 Explanations$ Math

4th Edition 路 809 pages

ISBN: 9781319113339

Chapter 11: Q.14 (page 693)

Use your calculator鈥檚 RandInt function to generate 200 digits from $0$ to $9$ and store them in a list.
(a) State appropriate hypotheses for a chi-square goodness-of-fit test to determine whether your calculator鈥檚 random number generator gives each digit an equal chance to be generated.
(b) Carry out the test. Report your observed counts, expected counts, chi-square statistic, P-value, and your conclusion

Short Answer

Expert verified

(a)The null and alternative hypotheses are:

$H_{0} : p_{1} = p_{2} = p_{3} = p_{4} = 鈥� 鈥� 鈥� 鈥� . . = p_{10} = 0.10$

$H_{a}$ : At least one of the $p_{i}$ is incorrect

(b)There is insufficient evidence to reject the claim regarding the random digits claim.

Step by step solution

Part(a) Step 1: Given information

Use the RandInt function on your calculator to produce $200$ digits from $0$ to $9$ and save them in a list.

To establish whether your calculator's random number generator provides each digit an equal probability of being created, we'll need to state relevant hypotheses for a chi-square goodness-of-fit test.

Part(a) Step 2: Explanation

There are $10$ random digits, with the fraction of each digit chosen being the same.

The formula used here is

$蠂^{2} = 鈭� \frac{(O - E)^{2}}{E}$

The probability of choosing one digit is:

$p = \frac{1}{10} = 0.10$

Part (b) Step 1: Given information

Given in the question that, Use your calculator鈥檚 RandInt function to generate $200$ digits from $0$ to $9$ and store them in a list.

We need to report our observed counts, expected counts, chi-square statistic, P-value, and your conclusion After the test.

Part(b) Step 2: Explanation

The following is how the test statistic is calculated:

Observed value	Expected value	(O-E)	${(O - E)}^{2}$	$\frac{{(O - E)}^{2}}{E}$
$22$	$20$	$2$	$4$	$0.2$
$21$	$20$	$1$	$1$	$0.05$
$24$	$20$	$4$	$16$	$0.8$
$15$	$20$	$- 5$	$25$	$1.25$
$17$	$20$	$- 3$	$9$	$0.45$
$20$	$20$	$0$	$0$	$0$
$19$	$20$	$- 1$	$1$	$0.05$
$21$	$20$	$1$	$1$	$0.05$
$18$	$20$	$- 2$	$4$	$0.2$
$23$	$20$	$3$	$9$	0.45
				$鈭� (O - E) E 2 = 3.5$

Part(b) Step 3: Test statistic

The test statistic equation will be:

$蠂^{2} = 鈭� \frac{(O - E)^{2}}{E}$

The degree of freedom can be calculated in a way that:

$D e g r e e o f f r e e d o m = N u m b e r o f c a t e g o r i e s - 1 = 10 - 1 = 9$

At $9$ degrees of freedom, the $p$ -value will be $0.941$ byusing the chi-square table. The $p$ -value exceeds the significance level. In this scenario, the null hypothesis is not rejected.