Question

Much concern has been expressed 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 up take of radio-labeled amino acid (in dpm, disintegrations per minute) 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 know that the mean rate of uptake for cultures without nitrates is 8000. Do the data suggest that the addition of nitrates results in a decrease in the mean rate of uptake? Test the appropriate hypotheses using a significance level of .10.
Please use all 9 steps to answer this question with exact numbers in the question. Thank you!!
1. Describe the population characteristic about which hypotheses are to be tested.
2. State the null hypothesis H0.
3. State the alternative hypothesis Ha.
4. Choose the significance level, a, for the test. Note: Many times a will be specified in the problem, so just state it here. If it is not given, you are free to choose an appropriate significance level like a = .05
5. Check to make sure that any conditions/assumptions required for the test are reasonable. Note: Even if you believe that the assumptions have not been met, make a note of that and then say, "but we will proceed with the test anyway".
6. Clearly display the test statistic to be used and calculate the observed value of the test statistic.
7. Determine the P-value associated with the observed value of the test statistic.
8. State the conclusion (reject H0 or fail to reject H0) with reference to the level of significance.
9. State a final conclusion in the context of the problem.

Answer 1

Answer:

1) μ = 8000

2)  H0:  μ ≥8000

3)   Ha: μ  < 8000

4)  α= 0.10

5)  If the sample size is at least 15 a student's t test can be used.And we assume that the variances of the two independent groups are almost equal.

6) t= x`- μ/ s.d/ √n  = -0.816

7) The P- value = 0.214083.

8) The critical region is [-∞, -1.345]

9) Fail to Reject the Null Hypothesis

Step-by-step explanation:

1) The population characteristic is the population mean =μ = 8000

2) The claim is that the  addition of nitrates results in a decrease in the mean rate of uptake i.e μ < 8000

The null hypothesis is that  μ ≥ 8000

2) Null hypothesis: H0:  μ ≥8000

3) Alternative hypothesis:  Ha: μ  < 8000

4) The significance level alpha = α= 0.10

5)  If the sample size is at least 15 a student's t test can be used.And we assume that the variances of the two independent groups are almost equal.

Calculating mean and standard deviation using a calculator

Sample Size, n: 15

Mean= ∑ x/n = x`=: 7788.8

∑ x/n= 116832/15

St Dev, s.d(n-1) : 1002.4307

6) Test Statistic: student's t Test with d.f = n-1 = 14

t= x`- μ/ s.d/ √n  

t= 7788.8-8000/1002.4307/√15

t= -0.816

7) The P- value for the observed t value  is 0.214083.

The result is not significant at p < 0.10. H0 is accepted.

8) The Critical value for one sided t test at α =0.1 with 14 d.f is  less than t=  -1.3406

Hence the critical region is [-∞, -1.345]

Since the calculated t -0.816  value does not fall in the critical region  [-∞, -1.345] there is not sufficient evidence to reject H0

9) Fail to Reject the Null Hypothesis

Sample does not provide enough evidence to support the null hypothesis and conclude that  the data suggest that the addition of nitrates do not  result in a decrease in the mean rate of uptake. The claim is rejected.