A researcher wanted to determine if carpeted rooms contain more bacteria than on

Question

A researcher wanted to determine if carpeted rooms contain more bacteria than on carpeted rooms the table shows the results for the number of bacteria per cubic foot for both types of rooms.
Full data set:
Carpeted 10.1, 13, 15.2 9.8, 11.8, 11.4 12.7, 6.7
Uncarpeted 8.6, 13.3, 8.8 7.5, 7.3, 7.5 9.5, 10
Determine the P-value for this hypothesis test.
P-value =________ A researcher wanted to determine if carpeted rooms contain more bacteria than on carpeted rooms the table shows the results for the number of bacteria per cubic foot for both types of rooms.
Full data set:
Carpeted 10.1, 13, 15.2 9.8, 11.8, 11.4 12.7, 6.7
Uncarpeted 8.6, 13.3, 8.8 7.5, 7.3, 7.5 9.5, 10
Determine the P-value for this hypothesis test.
P-value =________
Full data set:
Carpeted 10.1, 13, 15.2 9.8, 11.8, 11.4 12.7, 6.7
Uncarpeted 8.6, 13.3, 8.8 7.5, 7.3, 7.5 9.5, 10 Carpeted 10.1, 13, 15.2 9.8, 11.8, 11.4 12.7, 6.7
Uncarpeted 8.6, 13.3, 8.8 7.5, 7.3, 7.5 9.5, 10
Determine the P-value for this hypothesis test.
P-value =________

Explanation / Answer

Given that,
mean(x)=11.3375
standard deviation , s.d1=2.5399
number(n1)=8
y(mean)=9.0625
standard deviation, s.d2 =1.9748
number(n2)=8
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.89
since our test is right-tailed
reject Ho, if to > 1.89
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =11.3375-9.0625/sqrt((6.45109/8)+(3.89984/8))
to =2
| to | =2
critical value
the value of |t | with min (n1-1, n2-1) i.e 7 d.f is 1.89
we got |to| = 2.00003 & | t | = 1.89
make decision
hence value of | to | > | t | and here we reject Ho
p-value:right tail - Ha : ( p > 2 ) = 0.04281
hence value of p0.05 > 0.04281,here we reject Ho

ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 2
critical value: 1.89
decision: reject Ho
p-value: 0.04281

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