A researcher wanted to determine if carpeted rooms contain m

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. Determine whether carpeted rooms have more bacteria than uncarpeted rooms at the a = 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. H_0: Mu = Mu_2 H_1: Mu_1 Less than Mu_2 H_0: Mu_1 = Mu_2 H_1: Mu_1 not equal to Mu_2 H_0: Mu_1 = Mu_2 H_1: Mu_1 less than Mu_2 H_0: Mu_1 less than Mu_2 H_1: Mu_1 greater than Mu_2 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. Do not reject H_0. There is not significant evidence at the a = 0.01 level of significance to co

Solution

The reason it has population data provvide we use Z Test For Significance of Difference of Means

Set Up Hypothesis
Null, carpeted has similar bacteria as uncarpeted Ho: u1 = u2
Alternate , carpeted has more bacteria as uncarpeted H1: u1 > u2
Test Statistic
X(Mean)=10.575
Standard Deviation(s.d1)=1.955
Number(n1)=8
Y(Mean)=10.0875
Standard Deviation(s.d2)=2.2351
Number(n2)=8
we use Test Statistic (Z) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
Zo=10.575-10.0875/Sqrt((3.82203/8)+(4.99567/8))
Zo =0.46
| Zo | =0.46
Critical Value
The Value of |Z | at LOS 0.01% is 2.326
We got |Zo | =0.464 & | Z | =2.326
Make Decision
Hence Value of | Zo | < | Z | and Here we Do not Reject Ho
P-Value: Right Tail -Ha : ( P > 0.46 ) = 0.3212
Hence Value of P0.01 < 0.3212,Here We Do not Reject Ho

ANSWERS:
1. Ho: u1 = u2 , H1: u1 > u2
2. P-Value: 0.3212
3. Option A