Thanks A researcher wanted to determine if carpeted rooms contain more bacteria
ID: 3335708 • Letter: T
Question
Thanks
A researcher wanted to determine if carpeted rooms contain more bacteria than uncarpeted rooms. The table shows the results for the number of bacteria per cubic foot for both types of rooms. Full data set Uncarpeted Carpeted 13.5 13.8 11.6 11.4 5.3 2.6 10.6 15.1 6.3 138 8.9 13.8 11.9 14.4 11.9 82 Determine whether carpeted rooms have more bacteria than uncarpeted rooms at the : 0.01 level of significance. Normal probability plots indicate that the data are approximately normal and boxplots indicate that there are no outliers State the null and alternative hypotheses. Let population 1 be carpeted rooms and population 2 be uncarpeted rooms.Explanation / Answer
Given that,
Carpeted data,
(13.5,13.8,11.6,10.6,15.1,6.3,11.9,14.4)
mean(x)=12.15
standard deviation , s.d1=2.8107
number(n1)=8
Uncarpeted data,
(11.4,5.3,12.6,13.8,8.9,13.8,11.9,8.2)
y(mean)=10.7375
standard deviation, s.d2 =3.0095
number(n2)=8
null, Ho: u1 <= u2
alternate,carpeted rooms have more bacteria than uncapeted rooms H1: u1 > u2
level of significance, = 0.01
from standard normal table,right tailed t /2 =2.998
since our test is right-tailed
reject Ho, if to > 2.998
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =12.15-10.7375/sqrt((7.90003/8)+(9.05709/8))
to =0.97
| to | =0.97
critical value
the value of |t | with min (n1-1, n2-1) i.e 7 d.f is 2.998
we got |to| = 0.97019 & | t | = 2.998
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:right tail - Ha : ( p > 0.9702 ) = 0.18214
hence value of p0.01 < 0.18214,here we do not reject Ho
ANSWERS
---------------
a.
Ho: u1 = u2
H1: u1 > u2
b.
test statistic: 0.97
critical value: 2.998
decision: do not reject Ho
p-value: 0.18214
c.
Do not reject H0: There is not significamce evidence at alpha=0.01 level of significance that carpeted rooms have more bacteria than uncapeted rooms
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