Question

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4) An educator believes that new reading activities in the classroom will help elementary school pupils improve their reading ability. She arranges for a third grade class of 10 students to follow these activities for an 8-week period. A control classroom of 12 third graders follows the same curriculum without the activities. At the end of the 8 weeks, all students are given the Degree of Reading Power (DRP) test, which measures the aspects of reading ability that the treatment is designed to improve. The table below represents the results of this test (assume the distributions of the treatment and control groups are normally distributed). See Data Set for values. 46 43 Treatment 40 Control 62 43 49 59 55 71 34 52 57 41 18 5342 57 45 X 35 X 41 45 42 Based on the evidence provided, does appear that these new reading activities actually improve scoring on the DRP at the 5% significance level? YES NO Please show the four steps of your analysis in the space provided below...

Answer 1

4.

Given that,
mean(x)=50.9
standard deviation , s.d1=10.0604
number(n1)=10
y(mean)=43.4167
standard deviation, s.d2 =11.3735
number(n2)=12
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.833
since our test is right-tailed
reject Ho, if to > 1.833
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =50.9-43.4167/sqrt((101.21165/10)+(129.3565/12))
to =1.637
| to | =1.637
critical value
the value of |t α| with min (n1-1, n2-1) i.e 9 d.f is 1.833
we got |to| = 1.63686 & | t α | = 1.833
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value:right tail - Ha : ( p > 1.6369 ) = 0.06804
hence value of p0.05 < 0.06804,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 1.637
critical value: 1.833
decision: do not reject Ho
p-value: 0.06804
we do not have enough evidence to support the claim that the new reading activites actually improve score on the DRP.