At Alberton High School, 19 college-bound seniors all take a standardized colleg

Question

At Alberton High School, 19 college-bound seniors all take a standardized college admissions test. Their mean score is 51.1 with a standard deviation of 8.7.

Meanwhile, at Brownville High School, 12 college-bound seniors all take the same standardized college admissions test. Their mean score is 63.6 with a standard deviation of 7.4.

(a) Find the standard error of the difference between the mean score at Alberton and the mean score at Brownville.
2.92353
2.7971
10.4029
2.9765

(b) Find the 90% confidence interval of the difference of the mean scores.
(-17.7503,-7.24969)
(-18.6394,-6.36059)
(-17.5577,-7.44229)
(-18.1424,-6.85759)

Now test the claim that the Alberton mean score was less than the Brownville mean score with significance level = 0.1.

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

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

(e) TT-score:
-4.27565
-2.945195
2.347376
-1.093994

(f) Interval containing the pp-value:
(0,0.005)
(0.005,0.01)
(0.01,0.025)
(0.025,0.05)
(0.05,0.1)
(0.1,1)

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

Explanation / Answer

Two-Sample T-Test and CI

Method

Equal variances are not assumed for this analysis.

Descriptive Statistics

Estimation for Difference

Test

a) standard error = sqrt(s1^2 / n1 + s2^2/n2) =sqrt( 8.7^2 /19 + 7.4^2 / 12) = 2.9235282

option A) 2.92353 is correct

b)   (-17.49, -7.51)

option C) is correct

c)

Null - The Alberton mean is equal to the Brownville mean

option D) is correct

d)Alternative hypothesis:

A) The Alberton mean is less than the Brownville mean is correct

e) T-score =-4.27565

option A) is correct

f)

p-value = 0.000

which is contained in (0,0.005)

g)

since p-value < 0.10 , we reject the null

There is sufficient data to support the claim.

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