Question

Chapter 9 The 94 students in a statistics class are categorized by gender and by the year in school. The numbers obtained are displayed belovw Year in school Gender Freshman Sophomore Junior Senior Graduate Total Male Female Total 31 63 94 17 17 19 23 24 13 24 Suppose we wish to test the null hypothesis that a student is twice likely to be a female than a male. Under the null: 1. a. What is the expected number of males? b. What is the expected number of females? c. What is the null and alternative hypotheses? d. What is the test-statistic and its distribution? e. What is the p-value of the test? f. What is you conclusion? 2. Suppose we wish to test the null hypothesis that the students are equally distributed in different years. Under the null: a. What is the expected number of sophomores? b. What is the expected number of female sophomores? c. What is the null and alternative hypotheses? d. What is the test-statistic and its distribution? e. What is the p-value of the test? f. What is you conclusion? 3. Suppose we wish to test the null hypothesis that thereis no association between the year in school and gender. Under the null hypothesis a. What is the expected number of male sophomores? b. What is the expected number of female seniors? c. What is the null and alternative hypotheses? d. What is the test-statistic and its distribution? e. What is the p-value of the test? f. What is you conclusion? Suppose we wish to test the null hypothesis that the both male and female students are equally distributed in different years, and that a student is twice likely to be a female than a male. How this test is different from the test of no association? 4.

Answer 1

1)

(fo-fe)2/fe (31-31.333)2131.333 0.004 (63-62.667)162.667 0.002 Categories male female Observed 31 63 Expected 94 1/3-31.333 94*2/3-62.667 Sum- 94 0.005 94 The following null and alternative hypotheses need to be tested Ha: Some of the population proportions differ from the values stated in the null hypothesis This corresponds to a Chi-Square test for Goodness of Fit (2) Rejection Region Based on the information provided, the significance level is a0.05, the number of degrees of freedom is df 11, so then the rejection region for this test is R 3.841) The Chi-Squared statistic is computed as follows = 0.004 + 0.002 = 0.005 Since it is observed thatx0.005 not rejected 3.841, it is then concluded that the null hypothesis is

a)
Expected number of males = 31.333
b) 62.6667
c)
Ho: p1 = 1/3 , p2 = 2/3
Ha: not Ho

d)
TS = 0.005
chi-square distribution with 1 degree of freedom
e) p-value = chisq.dist.rt(0.005,1)
= 0.9436
f)
we fail to reject the null hypothesis

Please post rest questions again

a