2. Much concern has been expressed in recent years regarding the practice of usi

Question

2.     Much concern has been expressed in recent years regarding the practice of using nitrates as meat preservatives. In one study involving possible effects of these chemicals, bacteria cultures were grown in a medium containing nitrates. The rate of uptake of radio-labeled amino acid was then determined for each culture, yielding the following observations.

7251

6871

9632

6866

9094

5849

8957

7978

7064

7494

7883

8178

7523

8724

7468

Suppose that it is known that the true average uptake for cultures without nitrates is 8000. Do the data suggest that the addition of nitrates results in a decrease in the true average uptake? Test the appropriate hypotheses using a significance level of 0.10.

a.      State the name of the hypothesis test that you will be running.

b.       Define the parameter.

c.      State and check the assumptions.

d.       State the null and alternative hypotheses.

e.      State the significance level.

f.       Compute the value of the test statistic and report the degrees of freedom.

g.       Find the p-value.

h.      State whether you reject or fail to reject the null hypothesis and state your conclusion in the context of the problem.

i.      If the result is statistically significant, construct a confidence interval as a measure of effect size. If the result is not statistically significant, please say so.

7251

6871

9632

6866

9094

5849

8957

7978

7064

7494

7883

8178

7523

8724

7468

Explanation / Answer

Below are the null and alternate hypothesis
H0: mu >= 8000
H1: mu < 8000

Here the parameter under test is mean. In order to test this hypothesis, we use t-test. This is because we do not have information about the variance of the population. Also the sample size is less than 30.

From the given data we can compute,
mean   7788.8
std. dev.    1002.4308

Test statistics, t = (7788.8 - 8000)/(1002.4308/sqrt(15)) = -0.816

This is left tailed test, this means rejection region lies to the left of the graph.

Here degrees of freedom, df = 15 - 1 = 14
p-value = 0.2141

Here significance level = 0.05
As p-value is greater than significance level, we fail to reject the null hypothesis.

This means, the data does not suggest that the addition of nitrates results in a decrease in the true average uptake.

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.