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.

Carpeted   Uncarpeted
10.8    10.5
   6 8.4
9.8    6.5
8 6.7
14.7 11.7
9.2    5.5
15.6 13.3
9.9 10.6

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

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

Answer 1

We want to whether carpeted rooms have more bacteria than uncarpeted rooms at α=0.01 level of significance.

We have given that Normal probability plots indicate that the data are approximately normal and boxplots indicate that there are no outliers.

So we conduct two sample t test with pooled (equal variance) variance because there is no large difference in variances of the two groups.

Let population 1 be carpeted rooms and population 2 be uncarpeted rooms.

µ₁ be a population mean of captured room and µ₂ be a population mean of uncaptured room.

H₀ : µ₁ = µ₂

H₁ : µ₁ > µ₂

You will get the following output in Excel

t-Test: Two-Sample Assuming Equal Variances
	Variable 1	Variable 2
Mean	10.5	9.15
Variance	10.39714286	7.794286
Observations	8	8
Pooled Variance	9.095714286
Hypothesized Mean Difference	0
df	14
t Stat	0.895252122
P(T<=t) one-tail	0.192893093
t Critical one-tail	2.624494068
P(T<=t) two-tail	0.385786186
t Critical two-tail	2.976842734

From the above output,

Test statistics = 0.895

Since this test is one sided.

P value = 0.1929

Since p value = 0.1929 > 0.01 therefore we failed to reject null hypothesis.

Conclusion: There is no sufficient evidence to claim than carpeted rooms have more bacteria than uncarpeted rooms at α=0.01 level of significance.