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(2 points) We want to test whether or not more students are absent on Friday afternoon classes than on Wednesday afternoon classes. In a random sample of 300 students with Friday afternoon classes, 62 missed the class. In a different random sample of 300 students with Wednesday afternoon classes, 20 missed the class. The table below summarizes this information. The standard error (SE) is given to save calculation time if you are not using software. Class Day Friday total # of absences (x) 62 20 total # of students (n) 300 300 proportion p xin 0.206666666666667 Standard Error: SE 0.0280462580853951 Test the claim that the absentee rate on all Friday afternoon classes is greater than the absentee rate on all Wednesday afternoon classes. Test this claim at the 0.05 significance level. (a) Find the test statistic (to three decimal values) (b) Find the p-value (if smaller than 1.0e-5, write 0) (c) Is there sufficient data to support the claim at the 0.05 significance level? Yes No (d) Is there sufficient data to support the claim at the 0.01 significance level? Yes No Note: You can earn partial credit on this problem. Preview My Answers Submit Answers You have attempted this problem 0 times. You have 3 attempts remaining
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Answer #1

To test H_0:p_1 = p_2 against H_1:p_1 > p_2

Here,

i- 0.20667, n1300, p2 0.06667, n2-300

The test statistic can be written as

(Pi - /2) p1-p21-P2 which under H0 follows a standard normal distribution.

We reject H0 at 5% level of significance if P-value < 0.05

Now,

a) The value of the test statistic zobs 5.09859

b) P-value = P(Z > zobs) = P(Z > 5.09859) 1-1 = 0

c) Since p-value < 0.05, so we reject H0 at 5% level of significance and there is sufficient data to support the claim at 0.05 significance level.

d) Since p-value < 0.10, so we reject H0 at 5% level of significance and there is sufficient data to support the claim at 0.05 significance level.

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