Deliverable 5 - Hypothesis Tests for Two Samples WORD DOC/ EXCEL to show Calculations
ACL267
JPerez_Deliverable_5.xlsx1.xlsxInstructions - Read First
Instructions: The following worksheets describe two problems – the first problem is for independent samples and the second problem is for dependent samples. Your job is to demonstrate the solution to each scenario by showing how to work through each problem in detail. You are expected to explain all of the steps in your own words.
Independent Samples
| Low Lead Level | High Lead Level | ||
| n1 | 78 | n2 | 21 |
| 92.88 | 86.9 | ||
| s1 | 15.34 | s2 | 8.99 |
| Low Lead Level | High Lead Level | ||
| n1 | 78 | n2 | 21 |
| 92.88 | 86.9 | ||
| s1 | 15.34 | s2 | 8.99 |
| Critical Value: | |||
| Test Statistic: | |||
| p-value: | |||
| Critical Value: | 1.7247182429 | ||
| Test Statistic: | 2.2822693906 | ||
| p-value: | 0.0167782483 |
3. Make a decision about the null hypothesis and explain your reasoning, then make a conclusion about the claim in nontechnical terms. Evidence supports the claim that the test the claim that the mean IQ score of people with low lead levels is higher than the mean IQ score of people with high lead levels. The test statistic (2.28) would fall to the right of of the critical value (1.72) on a t number line.
Independent Samples A researcher conducted a test to learn the effect of lead levels in human bodies. He collected the IQ scores for a random sample of subjects with low lead levels in their blood and another random sample of subjects with high lead levels in their blood. The summary of finding is listed below. Use a 0.05 significance level to test the claim that the mean IQ score of people with low lead levels is higher than the mean IQ score of people with high lead levels. We do not know the values of the population standard deviations.
Dependent Samples
| Days of Release/Book | Phoenix | Prince | Difference | ||
| 1 | 44.2 | 58.2 | -14.0 | Mean of difference (d-bar) | -1.6 |
| 2 | 18.4 | 22.0 | -3.6 | Standard Deviation of Difference (S_d) | 4.5956622059 |
| 3 | 25.8 | 26.8 | -1.0 | ||
| 4 | 28.3 | 29.2 | -0.9 | ||
| 5 | 23.0 | 21.8 | 1.2 | Test Statistic | -1.0803178528 |
| 6 | 10.4 | 9.9 | 0.5 | Critical Value | 1.8331129327 |
| 7 | 9.1 | 9.5 | -0.4 | p-value | 0.8459507298 |
| 8 | 8.4 | 7.5 | 0.9 | ||
| 9 | 7.6 | 6.9 | 0.7 | ||
| 10 | 10.2 | 9.3 | 0.9 | ||
| Critical Value: | |||||
| Test Statistic: | |||||
| p-value: |
5. Calculate the critical value, the test statistic, and p-value. Show calculations below. Test Statistic = -1.08032 Critical Value = 1.833113 p-value = 0.845951 t.inv(1-a,df) 1-t.dist(t,df,true) t.inv(1-0.05,9) 1-t.dist(1.08, 9, true)
6. Make a decision about the null hypothesis and explain your reasoning, then make a conclusion about the claim in nontechnical terms. Since the p-value (0.85) is greater than alpha (0.05) we do not reject the Null hypothesis and evidence does not support the claim which of his favorite movies made more money.
Dependent Samples The Harry Potter books and movies made a lot of money. A fan wanted to learn which of his favorite movies made more money. He collected the amounts grossed in millions during the first few days of releases of the movies Harry Potter and the Half-Blood Prince and Harry Potter and the Order of the Phoenix. Use a 0.05 significance level to test his claim that the Prince movie did better at the box office. Use the p-value method to determine whether or not to reject the null hypothesis and state your conclusion.
