At Alberton High School, 49 college-bound seniors were sampled, and 7 were accep

Question

At Alberton High School, 49 college-bound seniors were sampled, and 7 were accepted into their top choice.

Meanwhile, at Brownville High School, 35 college-bound seniors were sampled, and 3 were accepted into their top choice.

(a) Find the standard error of the difference of the proportions of students who were accepted into their top choice between Alberton and Brownville.
0.0688333
0.252377
0.427652
0.498238

(b) Find the 90% confidence interval of the difference of proportions.
(-0.0560778,0.170364)
(-0.113564,0.227849)
(-0.0805237,0.194809)
(-0.105304,0.219589)

Now test the claim that the proportion of all college-bound seniors at Alberton was different than at Brownville with significance level = 0.1.

(c) Null hypothesis:
The Alberton proportion is less than the Brownville proportion
The Alberton proportion is greater than the Brownville proportion
The Alberton proportion is different than the Brownville proportion
The Alberton proportion is equal to the Brownville proportion

(d) Alternative hypothesis:
The Alberton proportion is less than the Brownville proportion
The Alberton proportion is greater than the Brownville proportion
The Alberton proportion is different than the Brownville proportion
The Alberton proportion is equal to the Brownville proportion

(e) ZZ-score:
0.830163
0.243729
-1.896849
2.532696

(f) pp-value:
0.406447
0.186801
0.037301
0.251801

(g) Result:
There is sufficient data to support the claim.
There is not sufficient data to support the claim.

Note: You can earn partial credit on this problem.

Explanation / Answer

a) Proportion of Alberton High School = 7/49 = 0.142857

Proportion of Brownville High School = 3/35 =0.085714

Standard error = Sqrt((0.142857*(1-0.142857)/49)+(0.085714*(1-0.085714)/35)) = 0.0688333

So, Option A is Correct

b) Margin of error = 1.645*0.0688333 = 0.113221

90% Interval

(0.142857-0.085714-0.0113221, 0.142857-0.085714-0.0113221)

= (-0.0560778, 0.170364)

So, Option A is Correct

c) Option A is Correct. The Alberton proportion is equal to the Brownville proportion

d) Option C is Correct.  The Alberton proportion is different than the Brownville proportion

e) Z score = (0.142857-0.085714)/0.068833 = 0.830163

So,Option A is Correct

f) P value = 0.406447

Option A is Correct

g) There is not sufficient data to support the claim.

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