Question

High School A and High School B both claim that they are superior. Last year School A had 354 students take an AP exam and 231 passed. School B had 684 students take the same exam and 512 passed.

Is this within natural variation? Or is this evidence that School B has a statistically significant higher passing rate? Report the p-value.

Answer 1

We are given that:

High School A and High School B both claim that they are superior.

School A: Total students 354; Passed 232 i.e., Failed = 354-232 = 122

Scholl B: Total Students 684; Passed 512 i.e. Failed = 684-512 = 172

We can solve this by the mathod of theory of attributes, or directly compairing the passing percentage for both the schools.

School	Passed	Fail	Total
A	231	123	354
B	512	172	684
Total	744	295	1039

Compare this with

Attributes	A		Total
B	(AB)	(B)	(B)
	(A)	()	()
Total	(A)	()	N

"If the attributes A and B are independent, the proportion of AB's in the population is equal to the product of the proportion of A's and B's in the population."

i.e.

Here

This is not a natural variatin,

we have i.e., negatively associated

2. is this evidence that School B has a statistically significant higher passing rate?

Null Hypothesis: H0: There is no significance difference between Scholl A and school B.

Alternative Hypothesis: Ha: School B is higher significant than school A or A is higher than B

we have chi-square test for 2X2 contingency table:

a	b
c	d

231	123
512	172

We have Tabulated Value for chi-square with 1 degree of freedom for 0.05% Level of significance is 3.841

Since the calculated value is much greater than the tabulated value, the value of chi-square is highly signifivant and Null Hypothesis is rejected. Hence we conclude that there is significance difference between both school

Now, Since the fraction of passing student in school A is

and , fraction of passing student in school B is

There for School B has a statistically significant higher passing rate.

p-value for Chi-square at 0.05% LOS and 1 Degree of freedom is 3.841'

p>0.05; Reject Null hypothesis.