A study is done by a community group in two neighboring colleges to determine which college graduates students with more math classes. College A samples 11 graduates. Their average is 4 math classes with a standard deviation of 1.5 math classes. College B samples 9 graduates. Their average is 3.5 math classes with a standard deviation of 1 math class. The community group believes that a student who graduates from college A has taken more math classes, on the average. Test at a 1% significance level. A. Is this a test of two means or two proportions? Are the populations dependent or independent?
B. Are the populations' standard deviations known or not?
C. Which distribution is to be used?
D. What is random variable?
E. Hypothesis
F. Test is two tailed or one tail
G. Test value=.
H. Use P-value method) to reach to the conclusion.

A) It is a test of two proportions and the populations are independent of each other

B) The population standard deviations are known : For college A the standard deviation is = 1.5 while for College B population the standard deviation is 1

C) The distribution to be used is 1 distribution

D) The Random variable is the size of the community group

E) Hypothesis : H0 : u1 = u2 . Ha : u1 > u2

F) Test is one tailed

G) Test Value  > significance level

F) when;  P value > significance level ( 0.01 ) we do accept Null hypothesis

   when ; P value < significance level ( 0.01 ) we reject Null hypothesis

Step-by-step explanation:

College A :

sample size = 11

x ( mean ) value = 4

std = 1.5

College B :

sample size = 9

x ( mean value ) = 3.5

std = 1

A) It is a test of two proportions and the populations are independent of each other

B) The population standard deviations are known : For college A the standard deviation is = 1.5 while for College B population the standard deviation is 1

C) The distribution to be used is 1 distribution

D) The Random variable is the size of the community group

E) Hypothesis : H0 : u1 = u2 . Ha : u1 > u2

F) Test is one tailed

G) Test Value  > significance level

F) when;  P value > significance level ( 0.01 ) we do accept Null hypothesis

   when ; P value < significance level ( 0.01 ) we reject Null hypothesis

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.