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Q 18.

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The Practice of Statistics for AP Examination
Found in: Page 718
The Practice of Statistics for AP Examination

The Practice of Statistics for AP Examination

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

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Short Answer

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 α=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 0.5 to 9.5 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)

H0 : p0 = p1 = p2 = p3 = p4 = p5 = p6 = p7 = p8 = p9 = 0.1 H1 : At least one of the Pi 's is incorrect

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

Part (c) 0.05

Part (d) P (At least one of the 25 make type I error)= 0.7226

See the step by step solution

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 =110 = 0.1

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

H0 : 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.

H1 : At least one of the Pi 's is incorrect

Part (b) Step 1: Calculation

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 χ2 = 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 χ2-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 Fail to reject H0

Part (c) Step 1: Calculation

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

Part (d) Step 1: Calculation

Multiplication rule

P (A B) = P (A and B) = P (A) × P (B)

Complement rule:

P (Ac ) = P ( not A) = 1 P (A)

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

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

= P (No Type I error) × P (No Type I error) × ... × P (No Type I error)25 repetitions= (P (No Type I error))2= 0.95 25= 0.2774

Using the compliment rule

P (At least one of the 25 make type I error) = 1 P (None of the 25 make type I erro) = 1 0.2774= 0 7226

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