Assume that the assumptions and conditions for inference with a two-sample t-test are met. Test the indicated claim about the means of the two populations. State your conclusion. Researchers wanted to compare the effectiveness of a water softener used with a filtering process to that of a water softener used without filtering. Ninety locations were randomly divided into two groups of equal size. Group A locations used a water softener and the filtering process, while group B used only the water softener. At the end of three months, water samples were tested at each location for softness level. (Water softness was measured on a scale of 1 to 5, with 5 being the softest water.) The results were as follows.
Group A (water softener and filtering)
xbar1 = 2.1
s1 = 0.7
Group B ( water softener only)
xbar2 = 1.7
s2 = 0.4
Using a 5% significance level, determine whether there is a difference between the two types of treatments.
H0: μ1 - μ2 = 0
Ha: μ1 - μ2 ≠ 0
Test statistic t = 3.328, P-value = 0.0014, DF = 69.97
A. Since the P-value is less than 0.05, we accept the null hypothesis and conclude that there is no difference in the softness level achieved from the two processes.
B. Since the P-value is less than 0.05, we are unable to reject the null hypothesis; there is insufficient evidence to conclude that there is a difference in the softness level achieved by the two processes.
C. Cannot draw a conclusion from the information given.
D. Since the P-value is less than 0.05, we reject the null hypothesis and conclude that there is a difference in the softness level achieved by the two processes.
E. Since the P-value is less than 0.05, we reject the alternative hypothesis and conclude that there is no significant difference in the softness level achieved by the two processes.

Respuesta :

Answer:

D. Since the P-value is less than 0.05, we reject the null hypothesis and conclude that there is a difference in the softness level achieved by the two processes.

Step-by-step explanation:

H0: μ1 - μ2 = 0

Ha: μ1 - μ2 ≠ 0

Test statistic t = 3.328, P-value = 0.0014, DF = 69.97

The Critical value for df  = 69.97 is t ( ∝/2) (df): ±1.9945

The  calculated t value falls in the critical region so H0 is rejected

As P - value is less than significance level H0 is rejected .

It is a two tailed test and the P - value is0.0014 which is less than 0.05 the significance level.

This means the softness achieved by Process A  is considered to be not equal to the softness achieved by the Process-B .

Only choice D is correct.

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