Question

SAT coaching: A sample of 10 students took a class designed to improve their SAT math scores. Following are their scores before and after the class. Can you conclude that the mean increase in scores differs from 15 points? Let μ1 represent the mean score after the class and μd=μ1-μ2. Use the α=0.10 level and the P-value method with the table.

	Score
Before	408	378	467	470	473	443	459	426	493	382
After	407	396	488	489	473	448	473	428	525	382

1. State the appropriate null and alternate hypotheses.
2. Compute the test statistic. Round the answer to at least three decimal places.
3. Estimate the P-value. Identify the form of the interval based on the Critical Values for the Student's t Distribution Table.

d. Determine whether to reject H0

e. State a conclusion.

Answer 1

Given that,
null, H0: Ud = 0
alternate, H1: Ud != 0
level of significance, α = 0.1
from standard normal table, two tailed t α/2 =1.833
since our test is two-tailed
reject Ho, if to < -1.833 OR if to > 1.833
we use Test Statistic
to= d/ (S/√n)
where
value of S^2 = [ ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = -11
We have d = -11
pooled variance = calculate value of Sd= √S^2 = sqrt [ 2376-(-110^2/10 ] / 9 = 11.382
to = d/ (S/√n) = -3.056
critical Value
the value of |t α| with n-1 = 9 d.f is 1.833
we got |t o| = 3.056 & |t α| =1.833
make Decision
hence Value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -3.0561 ) = 0.0137
hence value of p0.1 > 0.0137,here we reject Ho
ANSWERS
---------------
a.
null, H0: Ud = 0
alternate, H1: Ud != 0
b.
test statistic: -3.056
critical value: reject Ho, if to < -1.833 OR if to > 1.833
d.
decision: Reject Ho
c.
p-value: 0.0137
e.
we have enough evidence to support the claim that class designed to improve their SAT math scores. Following are their scores before and after the class.