Researchers wanted to compare the effectiveness of a water softener used with a filtering process with 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, a water sample was tested at each location for its level of softness. (Water softness was measured on a scale of 1 to 5, with 5 being the softest water.) The results were as follows. x1-2.1 s1-0.7 x2-1.7 82 0.4 State the null and alternate hypothesis. Graph and shade the critical region. Find the critical value, the point estimate for the difference in population means given by these samples, and it's test statistic. Label these values and areas on your graph above. Find and explain the meaning of the P-value. Shade a graph showing the area equal to the p-value. Clearly state your initial and final conclusion

The researcher's objective is to compare the effectiveness of a water softener when used with a filtering process against its effectiveness when used without filtering.

To do so 90 locations were randomly divided into two equal groups.

Group A locations used the water softener with filtering.

Group B locations used the water softener without filtering.

At the end of three months, a water sample was taken of each location and its level of softness was registered (Scale 1 to 5, 5 represents the softest water)

X₁: Softness of water of a location from group A

n₁= 45 locations

X[bar]₁= 2.1

S₁= 0.7

X₂: Softness of water of a location from group B

n₂= 45 locations

X[bar]₂= 1.7

S₂= 0.4

To compare the effectiveness of the softener with and without a filtering process the parameter of interest is the difference between both population means:

Parameter: μ₁ - μ₂

The point estimation of the difference between the population means is the difference of the sample means: X[bar]₁ - X[bar]₂= 2.1-1.7= 0.4

Since the objective is to test if there is any difference with or without the filtering process, the hypothesis test to make is two-tailed:

H₀: μ₁ - μ₂ = 0

H₁: μ₁ - μ₂ ≠ 0

α: 0.05

Since there is no information about the distribution of both variables, you have to apply the central limit theorem and approximate the distribution of X[bar]₁ and X[bar]₂ to normal. Once both samples mean distribution is approximate to normal you can use the statistic:

[tex]Z= \frac{(X[bar]₁ - X[bar]₂)-(Mu_1-Mu_2)}{\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2} } }[/tex]

[tex]Z_{H_0}= \frac{(2.1-1.7)-0}{\sqrt{\frac{0.49}{45} +\frac{0.16}{45} } } = 3.3282[/tex]

As said before, this test is two-tailed, so you will have two critical values:

Critical value 1: [tex]Z_{\alpha /2}= Z_{0.025}= -1.95[/tex]

Critical value 2: [tex]Z_{1-\alpha /2}= Z_{0.975}= 1.965[/tex]

The p-value of this test is also two tailed, you can calculate it as:

P(Z≥3.33) + P(Z≤3.33)= (1 - P(Z≤3.33))+P(Z≤-3.33)= (1-0.999566)+0.000434= 0.000868

p-value: 0.000868

This value means that 0.0868% of the sample size 45 taken from this population will provide natural evidence that there is no difference between the population means of the effectiveness of the water softener used with and without a filtering process.

A little reminder, the p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).

Both using the critical value method and the p-value method the decision is to reject the null hypothesis. This means that with a 5% level of significance there is a difference between the true average of the effectiveness of the water softener used with a filtering process and the true average effectiveness of the water softener used without a filtering process.

I hope it helps!