Question

A standardized exam consists of three parts: math, writing, and critical reading. Sample data showing the math and writing scores for a sample of 12 students who took the exam follow.

Student	Math	Writing
1	540	468
2	432	380
3	528	463
4	574	612
5	448	420
6	502	526
7	480	430
8	499	459
9	610	609
10	572	541
11	390	335
12	593	613

(a)

Use a 0.05 level of significance and test for a difference between the population mean for the math scores and the population mean for the writing scores. (Use math score − writing score.)

Formulate the hypotheses.

H₀: μ_d ≤ 0

H_a: μ_d = 0

H₀: μ_d ≠ 0

H_a: μ_d = 0

H₀: μ_d > 0

H_a: μ_d ≤ 0

H₀: μ_d = 0

H_a: μ_d ≠ 0

H₀: μ_d ≤ 0

H_a: μ_d > 0

Calculate the test statistic. (Round your answer to three decimal places.)

Calculate the p-value. (Round your answer to four decimal places.)

p-value =

What is your conclusion?

Do not reject H₀. We can conclude that there is a significant difference between the population mean scores for the math test and the writing test. Reject H₀. We can conclude that there is a significant difference between the population mean scores for the math test and the writing test.      Do not reject H₀. We cannot conclude that there is a significant difference between the population mean scores for the math test and the writing test. Reject H₀. We cannot conclude that there is a significant difference between the population mean scores for the math test and the writing test.

(b)

What is the point estimate of the difference between the mean scores for the two tests? (Use math score − writing score.)

What are the estimates of the population mean scores for the two tests?

Math:?

Writing:?

Which test reports the higher mean score?

The math test reports a / lower or higher /mean score than the writing test.

Answer 1

X	Y	X-Y	(X-Y)^2
540	468	72	5184
432	380	52	2704
528	463	65	4225
574	612	-38	1444
448	420	28	784
502	526	-24	576
480	430	50	2500
499	459	40	1600
610	609	1	1
572	541	31	961
390	335	55	3025
593	613	-20	400
		312	23404

Given that,
null, H0: Ud = 0
alternate, H1: Ud != 0
level of significance,alpha = 0.05
from standard normal table, two tailed talpha/2 =2.201
since our test is two-tailed
reject Ho, if to < -2.201 OR if to > 2.201
we use Test Statistic
to= d/ (S/√n)
where
value of S^2 = [ ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 26
We have d = 26
pooled standard deviation = calculate value of Sd= √S^2 = sqrt [ 23404-(312^2/12 ] / 11 = 37.2851
to = d/ (S/√n) = 2.4156
critical Value
the value of |talpha| with n-1 = 11 d.f is 2.201
we got |t o| = 2.4156 & |talpha| =2.201
make Decision
hence Value of | to | > | talpha| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 2.4156 ) = 0.0343
hence value of p0.05 > 0.0343,here we reject Ho
------------------------------------------------------------------------------
(a)
null, H0: Ud = 0
alternate, H1: Ud != 0
test statistic: 2.412
critical value: reject Ho, if to < -2.201 OR if to > 2.201
decision: reject H0. We can conclude that there is a significant difference between the population mean scores for the math test and the writing test
p-value: 0.0343
(b)
Confidence Interval
CI = d ± t a/2 * (Sd/ Sqrt(n))
Where,
d = ∑ di/n
Sd = Sqrt( ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
d = ( ∑ di/n ) =312/12=26
Pooled Sd( Sd )= Sqrt [ 23404- (312^2/12 ] / 11 = 37.285
Confidence Interval = [ 26 ± t a/2 ( 21.527/ Sqrt ( 12) ) ]
= [ 26 - 2.201 * (10.763) , 26 + 2.201 * (10.763) ]
= [ 2.31 , 49.69 ]
difference between the mean scores for the two tests = point estimate = d = 312
by looking at confidence onterval we conclude taht math performence is better