Question

To investigate water quality, in early September 2016, the Ohio Department of Health took water samples at 24 beaches on Lake Erie in Erie County. Those samples were tested for fecal coliform, which is the E.coli bacteria found in human and animal feces. An unsafe level of fecal coliform means there is a higher chance that disease‑causing bacteria are present and more risk that a swimmer will become ill if she or he should accidentally ingest some of the water. Ohio considers it unsafe for swimming if a 100 ‑milliliter sample (about 3.4 ounces) of water contains more than 400 coliform bacteria. The E. coli levels found by the laboratories are shown in the table.

18.7 579.4 1986.3 517.2 98.7 45.7 124.6 201.4
19.9 83.6 365.4 307.6 285.1 152.9 18.7 151.5
365.4 238.2 209.8 290.9 137.6 1046.2 127.4 224.7
Take these water samples to be an SRS of the water in all swimming areas in Erie County. Let represent the mean E. colicounts for all possible 100‑mL samples taken from all swimming areas in Erie County. We test H0:=400 versus H:<400 because the researchers are interested in whether the average E. coli levels in these areas are safe.
(a) Find x⎯⎯⎯ , , and the statistic. (Enter your answers rounded to three decimal places)
x⎯⎯⎯=
=
=
Find the P-value . (Enter your answer rounded to four decimal places.)
P‑value=
Are these data good evidence that on average the E. coli levels in these swimming areas were safe?
a) The data gives us no conclusive evidence one way or the other.
b) There is not good evidence to conclude that swimming areas in Erie County have mean E. coli counts less than 400 bacteria per 100 mL.
c) There is good evidence to conclude that swimming areas in Erie County have mean E. coli counts less than 400 bacteria per 100mL.
(b) Use the software of your choice to make a graph of the data. The distribution is very skewed. Another method that gives P‑values without assuming any specific shape for the distribution gives a P‑value of 0.0043 to answer if the given data shows average E.coli levels were safe in the swimming areas.
How does the one‑sample test compare with this?
a) The one‑sample test gives a significantly lower P‑value.
b) Both methods give similar P‑values.
c) The one‑sample test gives a significantly higher P‑value.
Should the procedures be used with these data?
a) Due to symmetry and the absence of outliers, procedures should be used here.
b) Due to extreme skew and the presence of outliers, procedures should not be used here.
c) Due to symmetry and the absence of outliers, procedures should not be used here.
What does the P‑value from the method that does not assume any specific shape for the distribution indicate?
a) The method that does not assume a specific shape for the distribution provides very little evidence that these swimming areas are safe on average.
b) The method that does not assume a specific shape for the distribution provides very strong evidence that these swimming areas are safe on average.
c) The method that does not assume a specific shape for the distribution provides very strong evidence that these swimming areas are not safe on average.

Answer 1

The P-value represents the probability of observing a test statistic as extreme as the calculated t-value, given the null hypothesis is true. In this case, the P-value is 0.0043, indicating evidence to support the alternative hypothesis that the average E. coli levels are less than 400 bacteria per 100 mL.

Step-by-step explanation:

To find the mean of the E. coli levels in the water samples, we sum up all the values and divide it by the total number of samples. So, the mean x is calculated as follows:

x = (18.7 + 579.4 + 1986.3 + 517.2 + 98.7 + 45.7 + 124.6 + 201.4 + 19.9 + 83.6 + 365.4 + 307.6 + 285.1 + 152.9 + 18.7 + 151.5 + 365.4 + 238.2 + 209.8 + 290.9 + 137.6 + 1046.2 + 127.4 + 224.7) / 24

Now, to find the sample standard deviation s:

Calculate the squared deviations from the mean for each value.

Add up all the squared deviations.

Divide this sum by (n-1), where n is the number of samples.

Take the square root of the result.

Finally, to find the t-statistic, we use the formula:

t = (x - μ0) / (s / √n)

The P-value can be found using a t-table or statistical software, and it represents the probability of observing a test statistic as extreme as the calculated t-value, given the null hypothesis is true.

If the P-value is less than the significance level (usually 0.05), we reject the null hypothesis and conclude that there is evidence to support the alternative hypothesis, which states that the average E. coli levels are less than 400 bacteria per 100 mL.

Based on the given information, the P-value is 0.0043, which is smaller than 0.05. Therefore, we have evidence to conclude that the average E. coli levels in these swimming areas are less than 400 bacteria per 100 mL.