Question

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
		Carpeted				Uncarpeted
		7.9	7.2	11.6		4.6	6	5.4
		8.5	9.9	10.7		6.9	8.2	6.7
		8.6	13.7			10.8	8.5

Determine whether carpeted rooms have more bacteria than uncarpeted rooms at the α=0.05 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.

A. H0​: μ1=μ2 H1​: μ1≠μ2

B. H0​: μ1=μ2 H1​: μ1< μ2

C. H0​: μ1=μ2 H1​: μ1>μ2

D. H0​: μ1<μ2 H1​: μ1>μ2

Determine the​ P-value for this hypothesis test.

​P-value=__ (Round to three decimal places as​ needed.)

State the appropriate conclusion. Choose the correct answer below.

A.

Do not reject

H0.

There

is not

significant evidence at the

α=0.05

level of significance to conclude that carpeted rooms have more bacteria than uncarpeted rooms.

B.

Reject

H0.

There

is

significant evidence at the

α=0.05

level of significance to conclude that carpeted rooms have more bacteria than uncarpeted rooms.

C.

Reject

H0.

There

is not

significant evidence at the

α=0.05

level of significance to conclude that carpeted rooms have more bacteria than uncarpeted rooms.

D.Do not reject H0. There is significant evidence at the α=0.05 level of significance to conclude that carpeted rooms have more bacteria than uncarpeted rooms.

Answer 1

For Carpeted :

∑x = 78.1

∑x² = 795.21

n1 = 8

Mean , x̅1 = Ʃx/n = 78.1/8 = 9.7625

Standard deviation, s1 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(795.21-(78.1)²/8)/(8-1)] = 2.1633

For Uncarpeted :

∑x = 57.1

∑x² = 434.95

n2 = 8

Mean , x̅2 = Ʃx/n = 57.1/8 = 7.1375

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(434.95-(57.1)²/8)/(8-1)] = 1.9784-

-------

Null and Alternative hypothesis:

Ho : µ1 = µ2

H1 : µ1 > µ2

df = ((s1²/n1 + s2²/n2)²)/[(s1²/n1)²/(n1-1) + (s2²/n2)²/(n2-1) ] = 13.8897 = 14

Test statistic:

t = (x̅1 - x̅2)/√(s1²/n1 + s2²/n2) = (9.7625 - 7.1375)/√(2.1633²/8 + 1.9784²/8) = 2.5327

p-value :

p-value = T.DIST.RT(2.5327, 14) = 0.012

Conclusion:

B.

Reject H0. There is significant evidence at the α = 0.05 level of significance to conclude that carpeted rooms have more bacteria than uncarpeted rooms.