For the water quality study in the reading, about 60% of the macroinvertebrate water quality indicators have historically been associated with good health. The researchers suspect that water quality in the area has decreased. They conduct a hypothesis test using significance level a=0.05 with the following hypothesis. h0:p=0.6, ha:p<0.6 They collect new data and this time n=80 and x=38 indicate good health. a. The conditions for conducting a hypothesis test are satisfied. true or false? b. What is phat(round to three decimal places) c. Find the test statistic.(round to three decimal places) d. What is the p-value for this test? (round three decimal places) e. What is the decision for this test? Reject the null or fail to reject the null hypothesis? f. Which of the following best summarizes the conclusion of this test? a. there is insufficient evidence to conclude that water quality in the area has not decreased b. there is sufficient evidence to conclude that water quality in the area has not decreased. c. there is insufficient evedinece to conclude that water quality in the area has decreased d. there is sufficient evidence to conclude the water quality in the area has decreased.

e) So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis

f) d. there is sufficient evidence to conclude the water quality in the area has decreased.

Reason : We reject the null hypothesis

Step-by-step explanation:

Data given and notation

n=80 represent the random sample taken

X=38 represent the people with good health

[tex]\hat p=\frac{38}{80}=0.475[/tex] estimated proportion of people with good health

[tex]p_o=0.6[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)

Part a

For this case we can check if we can use the normal approximation with these two conditions:

1) np = 80*0.475 = 38>10

2) n(1-p)=80*(1-0.475)= 42>10

So then we have the conditions to apply the z test

Part b

[tex]\hat p=\frac{38}{80}=0.475[/tex] estimated proportion of people with good health

Part c

We need to conduct a hypothesis in order to test the claim that the true proportion is loer than 0.6.:

Null hypothesis:[tex]p \geq 0.6[/tex]

Alternative hypothesis:[tex]p < 0.6[/tex]

When we conduct a proportion test we need to use the z statisitc, and the is given by:

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Part c: Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

[tex]z=\frac{0.475 -0.6}{\sqrt{\frac{0.6(1-0.6)}{80}}}=-2.28[/tex]

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

Part d

The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

[tex]p_v =P(z<-2.28)=0.011[/tex]

Part e

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis

Part f

d. there is sufficient evidence to conclude the water quality in the area has decreased.

Reason : We reject the null hypothesis