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Q 18.

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Found in: Page 718

### The Practice of Statistics for AP Examination

Book edition 6th
Author(s) Daren Starnes, Josh Tabor
Pages 837 pages
ISBN 9781319113339

# Is your random number generator working? Use your calculator’s RandInt function to generate $200$ digits from $0$ to $9$ and store them in a list.a. State appropriate hypotheses for a chi-square test for goodness of fit to determine whether your calculator’s random number generator gives each digit an equal chance of being generated.b. Carry out a test at the $\alpha =0.05$ significance level. Hint: To obtain the observedcounts, make a histogram of the list containing the $200$ random digits, and use the trace feature to see how many of each digit were generated. You may have to adjust your window to go from $-0.5to9.5$ with an increment of $1$c. Assuming that a student’s calculator is working properly, what is the probability that the student will make a Type I error in part (b)?d. Suppose that $25$ students in an AP® Statistics class independently do this exercise for homework and that all of their calculators are working properly. Find the probability that at least one of them makes a Type I error.

Part (a)

${H}_{0}:p0=p1=p2=p3=p4=p5=p6=p7=p8=p9=0.1\phantom{\rule{0ex}{0ex}}{H}_{1}:Atleastoneofthe{P}_{i}\text{'}sisincorrect$

Part (b) There is not enough proof to reject the claim of random digits.

Part (c) $0.05$

Part (d) $P\left(Atleastoneofthe25maketypeIerror\right)=0.7226$

See the 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.

## Part (a) Step 2: Calculation

Each digit has the same number of different outcomes, and there are a total of ten digits. As a result, the likelihood of any arbitrary digit is $1$ in $10$:

$p=\frac{1}{10}=0.1$

The null hypothesis states that the category variable's given distribution is correct.

${H}_{0}:p0=p1=p2=p3=p4=p5=p6=p7=p8=p9=0.1$

The alternative hypothesis is that the categorical variable's indicated distribution is incorrect.

${H}_{1}:Atleastoneofthe{P}_{i}\text{'}sisincorrect$

## Part (b) Step 1: Calculation

Find the chi-square subtotals and observed frequencies.

The observed frequencies obtained by entering $randInt\left(0,9,200\right)$ into the calculator are represented by $O$

The test-statistic is $\chi 2=3.5$

The degree of freedom is $df=c-1=10-1=9$

The $P-$value is the chance of having the test statistic's value, or a value that is 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:

$P>0.05⇒Failtoreject{H}_{0}$

## Part (c) Step 1: Calculation

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

## Part (d) Step 1: Calculation

Multiplication rule

$P\left(A\cap B\right)=P\left(AandB\right)=P\left(A\right)×P\left(B\right)$

Complement rule:

$P\left({A}^{c}\right)=P\left(notA\right)=1-P\left(A\right)$

Result part (c):

$P\left(TypeIerror\right)=0.05$

Using the complement rule:

$P\left(NoTypeIerror\right)=1-P\left(TypeIerror\right)=1-0.05=0.95$

Assuming that the $25$ pupils are self-contained, the following multiplication rule can be used to separate events:

$=\underset{⏟25repetitions}{P\left(NoTypeIerror\right)×P\left(NoTypeIerror\right)×...×P\left(NoTypeIerror\right)}$$={\left(P\left(NoTypeIerror\right)\right)}^{2}\phantom{\rule{0ex}{0ex}}={0.95}^{25}\phantom{\rule{0ex}{0ex}}=0.2774$

Using the compliment rule

$P\left(Atleastoneofthe25maketypeIerror\right)\phantom{\rule{0ex}{0ex}}=1-P\left(Noneofthe25maketypeIerro\right)\phantom{\rule{0ex}{0ex}}=1-0.2774\phantom{\rule{0ex}{0ex}}=07226$