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 observed

counts, 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.

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

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.

$H_1:Atleastoneofthe{P}_i\text{'}sisincorrect$

Find the chi-square subtotals and observed frequencies.

The observed frequencies obtained by entering $randInt(0,9,200)$ 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:

$P>0.05\Rightarrow Failtoreject{H}_0$

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

Multiplication rule

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

Complement rule:

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

Result part (c):

$P(TypeIerror)=0.05$

Using the complement rule:

$P(NoTypeIerror)=1-P(TypeIerror)=1-0.05=0.95$

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

$=\underset{⏟25repetitions}{P(NoTypeIerror)\times P(NoTypeIerror)\times ...\times P(NoTypeIerror)}$$$\begin{gathered} ={(P(NoTypeIerror\left)\right)}^2 \\ ={0.95}^{25} \\ =0.2774 \end{gathered}$$

Using the compliment rule