Question

Chapter 4, Section 2, Exercise 079 What is your lucky number? Thirty students are asked to choose a random number between and inclusive, to create a data set of n = digits. If the numbers are truly random, we would expect about 3 zeros 3 ones, 3 twos, and so on. If the data set includes 8 sevens, how unusual is that? If we look exclusively at the number of sevens, we expect the proportion of sevens to be 0.1 (since there are 10 possible numbers) and the number of sevens to be 3 in a sample size of 30. we are testing Ho :p0.1 vs Ha :P#0.1 where p is the proportion of sevens. We can generate the randomization distribution by generating 1000 sets of 30 random digits and recording X = the number of sevens in each simulated sample. See the figure below. 250 229 227 200 183 150 135 106 亡100 50 50-48 15 6 Sevens (a) Notice that this randomization distribution is not symmetric. This is a two-tailed test, so we need to consider both "tails". How far is X = 9 from the expected value of 3? what number would be equally far out on the other side? the expected value of 3 would be equally far out on the other side.

Answer 1

(b) Given The sample size, n 30 The total number of 7, x 8 The sample proportion of getting 7 = 0.2667 30 Stating hypothesis The null and alternative hypothesis is given by Но: р3D0.1 H p 01 Two tailed test) Significance level Let a 0.05 Test-statistics: For the single proportion test, always given by p-p sample mean proportion. Which is use z-test for one P1-р) 0.2667 0.1 0.1(1-0.1) 30 0.1667 0.099 V30 0.1667 0.0574 2.9019 _ Decision 0.00371|. (By using Atz 2.9019, the p-value of the respective two-tailed test NORM.DISTO function in Excel) Here, p-value level of significance. Hence, we reject the null hypothesis

Conclusion: At 0.05 level of significance, there is sufficient evidence to conclude that, the proportion of getting 7 is not equals to 0.1. (c) From the above-given figure: The sample size, n = 1,000 The number of zeros, x= 48 The proportion of getting a zero = 48 1,000 = 0.048 Stating hypothesis: The null and alternative hypothesis is given by: Ho: p=0.1 Hap<0.1 (One tailed test) Significance level: Let a=0.05 Test-statistics: For the single proportion test, always use z-test for one sample mean proportion. Which is given by: p-p p(1– p) 0.048-0.1 10.1(1-0.1) V 1.000 -0.052 0.099 V1,000 -0.052 0.00995 = -5.2262

Decision: At:=-5.2262, the p-value of the respective one-tailed test <0.0001). (By using NORM.DISTO function in Excel) Here, p-value < level of significance. Hence, we reject the null hypothesis. Conclusion: At 0.05 level of significance, there is sufficient evidence to conclude that, the proportion of getting is not less than 0.1.