Question

An education researcher would like to test whether 2nd graders retain or lose knowledge during the summer when they are presumably not in school. She asks nine 2nd graders to take a comprehension test at the end of the school year (May), and then asks those same students to come back after the summer (late August) to retake a different but equivalent test, to see if their level of comprehension has changed. Using the data below, test this hypothesis using an alpha level of .05. May August 90 100 65 80 78 92 50 60 89 90 92 98 75 70 90 96 65 87 What is the appropriate test? State the null hypothesis: State the alternative hypothesis: Find the critical value: Calculate the obtained statistic: Make a decision: What does your decision mean?   Use SPSS to perform this same analysis and confirm the same result. Paste your output below:

(12 pts) Below is a set of test 1 scores from our Statistics class. 7 of the scores come from females, and 7 come from males. Test the hypothesis that males and females scored differently on our first test, using an alpha level of .05. Male Female 65 99 90 78 87 43 98 56 46 72 61 90 70 100 What is the appropriate test? State the null hypothesis: State the alternative hypothesis: Find the critical value: Calculate the obtained statistic: Make a decision: What does your decision mean?

Answer 1

Given that,
mean(x)=77.1111
standard deviation , s.d1=14.7007
number(n1)=9
y(mean)=85.8889
standard deviation, s.d2 =13.5134
number(n2)=9
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.306
since our test is two-tailed
reject Ho, if to < -2.306 OR if to > 2.306
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =77.1111-85.8889/sqrt((216.11058/9)+(182.61198/9))
to =-1.319
| to | =1.319
critical value
the value of |t α| with min (n1-1, n2-1) i.e 8 d.f is 2.306
we got |to| = 1.31878 & | t α | = 2.306
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -1.3188 ) = 0.224
hence value of p0.05 < 0.224,here we do not reject Ho
ANSWERS
---------------
a.
T test for difference of means
b.
null, Ho: u1 = u2
c.
alternate, H1: u1 != u2
e.
test statistic: -1.319
d.
critical value: -2.306 , 2.306
f.
decision: do not reject Ho
p-value: 0.224
g.
we donot have enough evidence to support the claim that nine 2nd graders to take a comprehension test at the end of the school year (May), and then asks those same students to come back after the summer (late August) to retake a different but equivalent test,
to see if their level of comprehension has changed.
h.
Given that,
mean(x)=73.8571
standard deviation , s.d1=18.4881
number(n1)=7
y(mean)=76.8571
standard deviation, s.d2 =21.6212
number(n2)=7
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.447
since our test is two-tailed
reject Ho, if to < -2.447 OR if to > 2.447
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =73.8571-76.8571/sqrt((341.80984/7)+(467.47629/7))
to =-0.279
| to | =0.279
critical value
the value of |t α| with min (n1-1, n2-1) i.e 6 d.f is 2.447
we got |to| = 0.27901 & | t α | = 2.447
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -0.279 ) = 0.79
hence value of p0.05 < 0.79,here we do not reject Ho
ANSWERS
---------------
a.
T test for difference of means
b.
null, Ho: u1 = u2
c.
alternate, H1: u1 != u2
e.
test statistic: -0.279
d.
critical value: -2.447 , 2.447
f.
decision: do not reject Ho
p-value: 0.79
g.
we donot have enough evidence to support the claim that males and females scored differently on our first test,