Exercise #4
Quality control problem:
According to U.S. News & World Report, a laboratory in Atlanta is processing snake venom for use in medical research. Five processing methods are being tested to determine which is least likely to contaminate the processed venom. If a venom solution is tested as contaminated, it is marked with symbol “2B”. Each method produces 25 vials of venom each day. The vials from each processing method are then boxed for shipment to medical research facilities.
Assume for our purposes that scientists in the laboratory feel that the contamination is uniformly distributed, but others argue that it is not.
Conduct each test and state a conclusion using a significance level of 0.05.
processing method Observed
1 34
2 17
3 14
4 12
5 23
To conduct each test and state a conclusion using a significance level of 0.05, we will perform a hypothesis test for each processing method. Here are the steps for each test:
Step 1: Define the null and alternative hypotheses.
- Null hypothesis (H0): The processing method does not contaminate the venom solution.
- Alternative hypothesis (H1): The processing method contaminates the venom solution.
Step 2: Determine the test statistic.
- Since we are comparing observed values to expected values (assuming no contamination), we can use the chi-square goodness-of-fit test for each processing method.
Step 3: Set the significance level.
- The significance level is given as 0.05.
Step 4: Compute the expected values.
- Since each method produces 25 vials of venom each day, the expected value for each processing method is 25.
Step 5: Compute the test statistic.
- To calculate the chi-square test statistic, we use the formula:
X² = Σ [(Observed - Expected)² / Expected]
Here are the calculations for each processing method:
Processing method 1:
- Observed = 34
- Expected = 25
- X² = [(34 - 25)² / 25] = 3.44
Processing method 2:
- Observed = 17
- Expected = 25
- X² = [(17 - 25)² / 25] = 3.20
Processing method 3:
- Observed = 14
- Expected = 25
- X² = [(14 - 25)² / 25] = 4.40
Processing method 4:
- Observed = 12
- Expected = 25
- X² = [(12 - 25)² / 25] = 6.96
Processing method 5:
- Observed = 23
- Expected = 25
- X² = [(23 - 25)² / 25] = 0.32
Step 6: Calculate the p-value.
- We can compare the calculated test statistic to the chi-square distribution with 4 degrees of freedom (since we have 5 processing methods).
- Using a chi-square table or calculator, we find the p-values associated with the test statistics.
- For example, the p-value for X²=3.44 with 4 degrees of freedom is approximately 0.4869.
Step 7: Make a decision.
- Compare the p-value with the significance level of 0.05.
- If the p-value is less than 0.05, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
- For each processing method, we make the decision based on this criterion.
Conclusion for each processing method:
- Processing method 1: Since the p-value (0.4869) is greater than the significance level (0.05), we fail to reject the null hypothesis. There is not enough evidence to conclude that this method contaminates the venom solution.
- Processing method 2: Since the p-value (0.5345) is greater than the significance level (0.05), we fail to reject the null hypothesis. There is not enough evidence to conclude that this method contaminates the venom solution.
- Processing method 3: Since the p-value (0.4930) is greater than the significance level (0.05), we fail to reject the null hypothesis. There is not enough evidence to conclude that this method contaminates the venom solution.
- Processing method 4: Since the p-value (0.1425) is greater than the significance level (0.05), we fail to reject the null hypothesis. There is not enough evidence to conclude that this method contaminates the venom solution.
- Processing method 5: Since the p-value (0.9501) is greater than the significance level (0.05), we fail to reject the null hypothesis. There is not enough evidence to conclude that this method contaminates the venom solution.
To conduct the test and state a conclusion using a significance level of 0.05, we need to perform a hypothesis test.
Let's define our hypotheses:
Null Hypothesis (H0): The processing methods have no effect on the contamination of the venom. The mean contamination level for all processing methods is the same.
Alternative Hypothesis (Ha): The processing methods have an effect on the contamination of the venom. The mean contamination level for at least one processing method is different.
Now, let's calculate the test statistic using the observed data:
Step 1: Calculate the sample mean (x̄) and sample standard deviation (s) for each processing method.
Method 1:
x̄1 = 34
s1 = N/A (not provided)
Method 2:
x̄2 = 17
s2 = N/A (not provided)
Method 3:
x̄3 = 14
s3 = N/A (not provided)
Method 4:
x̄4 = 12
s4 = N/A (not provided)
Method 5:
x̄5 = 23
s5 = N/A (not provided)
Step 2: Calculate the overall sample mean (x̄) and overall sample standard deviation (s).
x̄ = (x̄1 + x̄2 + x̄3 + x̄4 + x̄5) / 5
= (34 + 17 + 14 + 12 + 23) / 5
= 100 / 5
= 20
s = N/A (not provided)
Step 3: Calculate the test statistic using the formula:
Test Statistic (F) = (Sum of Squares Between Groups) / (Sum of Squares Within Groups)
Where:
Sum of Squares Between Groups = [(n1*(x̄1 - x̄)^2) + (n2*(x̄2 - x̄)^2) + (n3*(x̄3 - x̄)^2) + (n4*(x̄4 - x̄)^2) + (n5*(x̄5 - x̄)^2)] / (k - 1)
Sum of Squares Within Groups = [(n1 - 1)*s1^2 + (n2 - 1)*s2^2 + (n3 - 1)*s3^2 + (n4 - 1)*s4^2 + (n5 - 1)*s5^2] / (N - k)
n1 = n2 = n3 = n4 = n5 = 25 (since each method produces 25 vials per day)
k = 5 (number of processing methods)
N = 5 * 25 = 125 (total number of vials)
Substituting the values:
F = [(25*(34 - 20)^2) + (25*(17 - 20)^2) + (25*(14 - 20)^2) + (25*(12 - 20)^2) + (25*(23 - 20)^2)] / (5 - 1)
/ [(24*(s1^2)) + (24*(s2^2)) + (24*(s3^2)) + (24*(s4^2)) + (24*(s5^2))] / (125 - 5)
Note: Since the sample standard deviation (s) is not provided for each method, we cannot calculate the test statistic. We would need this information to proceed with the hypothesis test.
Therefore, we cannot perform the hypothesis test or state a conclusion without the sample standard deviation for each processing method.