For a lottery to be successful, the public must have confidence in its fairness.

Question

For a lottery to be successful, the public must have confidence in its fairness. One of the lotteries in a state is a pick-3 lottery, where 3 random digits are drawn each day. A fair game depends on every value (0 to 9) being equally likely at each of the three positions. If not, then someone detecting a pattern could take advantage of that and beat the lottery. To investigate the randomness, we'll look at the data collected over a 32-week period. Although the winning numbers look like three-digit numbers, In fact, each digit is a randomly drawn numeral. We have 654 random digits in all. Are each of the digits from 0 to 9 equally likely? A table of the frequencies is shown to the right Complete parts a through e. a) Select the appropriate procedure. What kind of chi-square test would be appropriate? A. Chi square test for goodness-of-f t B. Chi square test for homogeneity C. Chi square test for independence D. A chi-square test would not be appropriate b) Check the assumptions. Are all of the assumptions and conditions satisfied? Select all that apply. A. No, the counted data condition is not satisfied. B. No, the Independence assumption is not satisfied. C. Yes. all of the assumptions and conditions are satisfied. D. No, the randomization condition Is not satisfied E. No, the expected cell frequency condition is not satisfied. c) State the hypotheses. H_0: The likelihood of drawing each numeral H_A: The likelihood of drawing each numeral Click to select your answer(s).

Explanation / Answer

Ans:a) Option A is correct.Chi square test for goodness of fit will be performed.

Critical chi square value at df=9 and significance level=0.05 is 16.92

The calculated value of Chi-Square goodness of fit test is compared with the critical value. If the calculated value of Chi-Square goodness of fit test is greater than the table value, we will reject the null hypothesis and conclude that there is a significant difference between the observed and the expected frequency.

Here,calculated chi square value is less than critical chi square value ,hence we can not ireject the null hypothesis i.e The likelihood of drawing each numeral is equally likely.

b)The Option C is correct.Yes,all of the assumptions and conditions are satisfied.

c)H0:The likelihood of drawing each numeral is equally likely.

HA:The likelihood of drawing each numeral is not equally likely.

Rererence:

The Chi Square Test is a statistical test which consists of three different types of analysis 1) Goodness of fit, 2) Test for Homogeneity, 3) Test of Independence.

The Test for Goodness of fit determines if the sample under analysis was drawn from a population that follows some specified distribution.

The Test for Homogeneity answers the proposition that several populations are homogeneous with respect to some characteristic.

The Test for independence (one of the most frequent uses of Chi Square) is for testing the null hypothesis that two criteria of classification, when applied to a population of subjects are independent. If they are not independent then there is an association between them.

Chi Square is the most popular discrete data hypothesis testing method.

Group Observed frequency Expected frequency (O-E)^2/E 0 63 65.4 0.0881 1 55 65.4 1.6538 2 66 65.4 0.0055 3 64 65.4 0.0300 4 75 65.4 1.4092 5 58 65.4 0.8373 6 71 65.4 0.4795 7 72 65.4 0.6661 8 68 65.4 0.1034 9 62 65.4 0.1768 5.4495

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