Question

The legislature of a southern state in the U.S. passed a rule, commonly called "no-pass, no-play", which prohibits a student who fails in any subject from participating in any extracurricular activity for six weeks. Data were collected for students involved in football, volleyball, cross country, and band for the first six-week grading period. Records were kept from last year and this year. The numbers of students is stored in column 1 and the proportions sidelined because of the rule are stored in column 2 of Table C, the first row being for last year and the second for this year.

29. What is the upper 90% confidence limit on the change in proportion of students sidelined because of failure?

30. What was the average (pooled) proportion sidelined?

31. Now use the pooled proportion to calculate the standard error of the difference between the two proportions. ) What is the value of the test statistic for testing the hypothesis that the proportion did not change (remember to divide by the standard error of the difference between the two proportions which was calculated using the pooled proportion)?

32. Was the superintendent of schools justified in saying, "We are very pleased with the improvement. It shows coaches and students are taking the rule seriously"? Answer 1 for yes or 0 for no.

DATA C:

273 0.29827189801646

256 . 0.28966880074687

Answer 1

Pooled sample proportion. Since the null hypothesis states that P1=P2, we use a pooled sample proportion (p) to compute the standard error of the sampling distribution.
p = (p1 * n1 + p2 * n2) / (n1 + n2)

here
p1 = 0.298 , n1 = 273
p2 = 0.289 , n2 = 256

putting the values and solving this we get

p = (p1 * n1 + p2 * n2) / (n1 + n2)
(0.298 * 273 + 0.289 * 256) / (273 + 256) = 0.293

where p1 is the sample proportion from population 1, p2 is the sample proportion from population 2, n1 is the size of sample 1, and n2 is the size of sample 2.
Standard error. Compute the standard error (SE) of the sampling distribution difference between two proportions.
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

sqrt{ 0.293 * ( 1 - 0.293 ) * [ (1/273) + (1/256) ] } =0.0396

where p is the pooled sample proportion, n1 is the size of sample 1, and n2 is the size of sample 2.
Test statistic. The test statistic is a z-score (z) defined by the following equation.
z = (p1 - p2) / SE

z = (0.298-0.289)/0.0396 = 0.2272

where p1 is the proportion from sample 1, p2 is the proportion from sample 2, and SE is the standard error of the sampling distribution.

now check the z table to calculate the p value
Two-tailed p-value: 2P(Z > |z|) = 0.8202682

as the p value is not less than 0.1(90%) , hence we fail to reject the null hypothesis and conclude that there is no difference between the 2 proportions