Question

2. (14 pts) Hardness of water from two different water treatment facilities is investigated. Observed water hardness (in ppm) for a random sample of faucets is as follows: Facility 1: 63 57 58 62 66 58 61 60 55 62 59 60 58 Facility 2: 69 65 59 62 61 57 59 60 60 62 61 66 68 66 Use α=0.01. (a) Assume σ1=σ2. Is there evidence to support the claim that the two facilities supply water of different hardness? (b) What is the p-value of this test? (c) Repeat question (a) but do not assume that σ1=σ2. (d) Find the p-value for part (c). (e) Compare the results from the two tests and discuss the reasons for differences. (f) Construct a 99% CI for the difference in water hardness for part (a). (e) Construct a 99% CI for the difference in the water hardness for part (b).

Answer 1

= 59.923, s1 = 2.9, n1 = 13

​​​​​​ = 62.5, s2 = 3.674, n2 = 14

A) H₀:

H₁:  

The pooled variance(s_p²) = ((n1 - 1)s1^2 + (n2 - 1)s2^2)/(n1 + n2 - 2) = (12 * (2.9)^2 + 13 * (3.674)^2)/(13 + 14 - 2) = 11.0559

The test statistic t = ()/sqrt(s_p²/n1 + s_p²/n2)

= (59.923 - 62.5)/sqrt(11.0559/13 + 11.0559/14) = -2.012

DF = 13 + 14 - 2 = 25

At alpha = 0.01, the critical values are t^* = +/- 2.787

Since the test statistic value is not less than the lower critical value (-2.012 > -2.787) , so we should not reject the null hypothesis.

So at alpha = 0.01, there is not sufficient evidence to support the claim that the two facilities supply water of different hardness.

B) P-value = 2 * P(T < -2.012)

= 2 * 0.0276 = 0.0552

C) The test statistic t = ()/sqrt(s1^2/n1 + s^2/n2)

= (59.923 - 62.5)/sqrt((2.9)^2/13 + (3.674)^2/14) = -2.03

Df = (s1^2/n1 + s^2/n2)^2/((s1^2/n1)^2/(n1 - 1) + (s^2/n2)^2/(n2 - 1))

= ((2.9)^2/13 + (3.674)^2/14)^2/(((2.9)^2/13)^2/12 + ((3.674)^2/14)^2/13) = 24

At alpha = 0.01, the critical values are t^* = +/- 2.797

Since the test statistic value is not less than the lower critical value (-2.03 > -2.797), so we should not reject the null hypothesis.

So at alpha = 0.01, there is not sufficient evidence to support the claim that the two facilities supply water of different hardness.

D) P-value = 2 * P(T < -2.03)

= 2 * 0.0268 = 0.0536

E) The result from the two tests are equal. In both the cases the null hypothesis is not rejected.

F) The 99% confidence interval is

() +/- t^* * sqrt(s_p²/n1 + s_p²/n2)

= (59.923 - 62.5) +/- 2.787 * sqrt(11.0559/13 + 11.0559/14)

= -2.577 +/- 3.569

= -6.146, 0.992

G )The 99% Confidence interval is

() +/- 2.797 * sqrt(s1^2/n1 + s^2/n2)

= (59.923 - 62.5) +/- 2.797 * sqrt((2.9)^2/13 + (3.674)^2/14)

= -2.577 +/- 3.55

= -6.127, 0.973