In: Statistics and Probability
1) A pharmaceutical company is about to launch a new manufacturing process in addition to the existing one. The quality control manager believes that the new method results in a different variation in the weights of the capsules. To verify the claim, the samples from each production line were obtained and the results are below (in mg):
Production Line 1:
101.4 | 102.5 | 100.2 | 98.3 | 100.3 |
101.2 | 98.8 | 98.2 | 99 | 99.6 |
101.1 | 101.9 | 102.3 | 100.3 | 98.4 |
99.3 | 102.2 |
(Note: The average and the standard deviation of the data are respectively 100.3 mg and 1.48 mg.)
Production Line 2:
97.6 | 99.9 | 98.6 | 97.9 | 97.2 |
97.7 | 98 | 98.6 | 97.9 | 99.5 |
99.6 | 99 | 99.5 | 99.2 |
(Note: The average and the standard deviation of the data are respectively 98.6 mg and 0.87 mg.)
Use a 5% significance level to test the claim that the standard deviation of the capsule weights in the production line 1 is greater than the standard deviation of the capsule weights in the production line 2.
Procedure: Select an answer Two means Z Hypothesis Test Two paired means Z Hypothesis Test Two proportions Z Hypothesis Test Two means T (pooled) Hypothesis Test Two means T (non-pooled) Hypothesis Test Two variances F Hypothesis Test
Assumptions: (select everything that applies)
Step 1. Hypotheses Set-Up:
H0:H0: Select an answer p₁-p₂ μ μ₁-μ₂ σ₁²/σ₂² = | , where Select an answer σ's are μ's are μ=μ₁-μ₂ is p's are the Select an answer population means difference between population means population standard deviations population proportions and the units are ? kg mg oz lbs |
Ha:Ha: Select an answer σ₁²/σ₂² μ p₁-p₂ μ₁-μ₂ ? > ≠ < | , and the test is Select an answer Two-Tail Right-Tail Left-Tail |
Step 2. The significance level α=α= %
Step 3. Compute the value of the test statistic: Select an answer z₀ t₀ f₀ χ²₀ = (Round the answer to 3 decimal places)
Step 4. Testing Procedure: (Round the answers to 3 decimal places)
CVA | PVA |
Provide the critical value(s) for the Rejection Region: | Compute the P-value of the test statistic: |
left CV is and right CV is | P-value is |
Step 5. Decision:
CVA | PVA |
Is the test statistic in the rejection region? | Is the P-value less than the significance level? |
? yes no | ? no yes |
Conclusion: Select an answer Do not reject the null hypothesis in favor of the alternative. Reject the null hypothesis in favor of the alternative.
Step 6. Interpretation:
At 5% significance level we Select an answer DO NOT DO have sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.
Question 2
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The proportion of people that own cats is 50%. A veterinarian believes that this proportion is larger than 50% and surveys 400 people. Test the veterinarian's claim at the α=0.01α=0.01 significance level.
Preliminary:
Test the claim:
Please also provide excel formulas that are used to calculate the Left and Right CVA, Z- score and P value
1. Option: Two variances F Hypothesis Test
Option: i. Normal populations; ii. Independent samples
Step 1. Hypotheses Set-Up:
(population standard deviations and unit mg).
Step 2. The significance level α=5%
Step 3. f0=2.1956/0.7644=2.872
Step 4:
Step 5. Decision: Yes, the test statistic is in the rejection region.
Yes, P-value is less than the significance level.
Conclusion: Reject the null hypothesis in favor of the alternative.
Step 6. Interpretation:
At 5% significance level we Select an answer DO have sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.
Conclusion: There is sufficient evidence to support the claim that the standard deviation of the capsule weights in the production line 1 is greater than the standard deviation of the capsule weights in the production line 2.
2. Yes, it is safe to assume that n≤0.05n≤0.05 of all subjects in the population.
n=400, p=0.5, np=200, n(1-p)=200
np(1-p)=400*0.5*0.5=100.
Test the claim: The null and alternative hypotheses are H0:p≤0.5, Ha:p>0.5
The test is right-tailed.
Since P-value<0.01, Reject the null hypothesis.
There is sufficient evidence that the proportion of people who own cats is larger.