An education researcher recruits seven middle-school students to test the effica
ID: 3319427 • Letter: A
Question
An education researcher recruits seven middle-school students to test the efficacy of an online spelling and grammar learning program. The students are asked to write a two-page essay before going through the program, and another two-page essay at the end. A software calculates the number of grammar and spelling errors in each essay. The results are in the table below.
Essay 1
Essay 2
26
27
28
23
19
15
19
20
19
13
20
11
17
8
a) Specify the hypotheses, both in words and in symbols, to test whether the results of the study are statistically significant. Specify the type of test you will use, and state all the necessary assumptions.
b) Determine the test value, the p-value, and summarize the conclusion of your test.
c) Determine a 95% confidence interval most appropriate for reporting the results of the data.
d) Explain how the results of your hypothesis test and your confidence interval support each other. Your explanation should include the confidence level and specific value(s) in the interval.
Essay 1
Essay 2
26
27
28
23
19
15
19
20
19
13
20
11
17
8
Explanation / Answer
a.
we use t test paired(before and after)
Given that,
null, H0: Ud = 0
alternate, H1: Ud > 0
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.447
since our test is two-tailed
reject Ho, if to < -2.447 OR if to > 2.447
we use Test Statistic
to= d/ (S/n)
where
value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 4.429
We have d = 4.429
pooled variance = calculate value of Sd= S^2 = sqrt [ 241-(31^2/7 ] / 6 = 4.158
to = d/ (S/n) = 2.818
critical Value
the value of |t | with n-1 = 6 d.f is 2.447
we got |t o| = 2.818 & |t | =2.447
make Decision
hence Value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 2.8185 ) = 0.0304
hence value of p0.05 > 0.0304,here we reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud > 0
b.
test statistic: 2.818
critical value: reject Ho, if to < -2.447 OR if to > 2.447
decision: Reject Ho
p-value: 0.0304
we have enough evidence to support the claim
c.
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 ) =31/7=4.429
Pooled Sd( Sd )= Sqrt [ 241- (31^2/7 ] / 6 = 4.158
Confidence Interval = [ 4.429 ± t a/2 ( 2.4/ Sqrt ( 7) ) ]
= [ 4.429 - 2.447 * (1.571) , 4.429 + 2.447 * (1.571) ]
= [ 0.583 , 8.274 ]
d.
yes,
the results of your hypothesis test and your confidence interval support each other
confidence interval = [ 0.583 , 8.274 ]
hypothesis test is reject the null hypothesis
X Y X-Y (X-Y)^2 26 27 -1 1 28 23 5 25 19 15 4 16 19 20 -1 1 19 13 6 36 20 11 9 81 17 8 9 81 0 0 0 0 0 0 31 241Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.