Question

An experiment was conducted to study the use of 95% ethanol versus 20% bleach as a disinfectant in removing contamination when culturing plant tissues. The experiment was repeated 15 times with each disinfectant, using cuttings of eggplant tissue. The observation reported was the number of uncontaminated eggplant cuttings after a 4-week storage.

Disinfectant 95% Ethanol 20% Bleach
Mean 3.74 4.86
Variance 2.78097 0.17142
n 15 15
Pooled variance 1.47620

1. Are you willing to assume that the underlying variances are equal?

a. Yes
b. No

2. Using the information from part (A), are you willing to conclude that there is a significant difference in the mean numbers of uncontaminated eggplants for the two disinfectants tested?

Answer 1

Answer:

a. Yes

Step-by-step explanation:

The term common variance means that each population has the same variance.

If σ²=σ₁²=σ₂² but unknown then the unbiased pooled or combined estimate of the common variance σ² is calculated.

In the given problem the pooled variance is calculated .

Therefore we assume that that the underlying variances are equal.

Using Student's t distribution

The given data is

Mean               3.74             4.86

Variance     2.78097       0.17142

n                       15                     15

Pooled variance 1.47620

t= x`1- x`2/ Sp √1/n1+ 1/n2

t = 3.74-4.86/ 1.47620 √1/15+ 1/15

t= -1.12/ 0.539

t= -2.0779

Taking the significance level ∝= 0.05 the

Value of the critical region is  t≤ - t∝/2  and  t≥  t∝/2 = ±2.048 for

Degrees of freedom = n1+n2- 2= 28

Hence the calculated value of t -2.0779 falls in the critical region therefore null hypothesis is rejected and  and it is concluded that the means are not equal.