For a certain drug, based on standards set by the United States Pharmacopeia (USP) - an official public standards-setting authority for all prescription and over-the-counter medicines and other health care products manufactured or sold in the United States, a standard deviation of capsule weights of less than 1.9 mg is acceptable. A sample of 34 capsules was taken and the weights are provided below:
| 120.5 | 122.4 | 118.5 | 119.7 | 119.8 |
| 119.9 | 122.1 | 117.3 | 119.4 | 116.2 |
| 123.4 | 119.7 | 120.2 | 122.3 | 119.7 |
| 120.7 | 118.1 | 122.1 | 120.1 | 120.9 |
| 120.9 | 121.8 | 119.4 | 123 | 119.5 |
| 116.6 | 118.8 | 115.7 | 122.7 | 119.6 |
| 123.4 | 120.9 | 123.2 | 117.7 |
(Note: The average and the standard deviation of the data are respectively 120.2 g and 2.06 g.)
At 1% significance level, test the claim that the standard deviation of capsule weights of the drug is greater than 1.9 g.
Procedure: Select an answer One variance χ² Hypothesis Test One proportion Z Hypothesis Test One mean Z Hypothesis Test One mean T Hypothesis Test
Assumptions: (select everything that applies)
Step 1. Hypotheses Set-Up:
| H0:H0: Select an answer σ² p μ = | , where ? p σ μ is the Select an answer population proportion population standard deviation population mean and the units are ? mg g lbs kg |
| Ha:Ha: Select an answer p μ σ² ? > < ≠ | , 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 χ²₀ t₀ f₀ = (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 | ? yes no |
Conclusion: Select an answer Reject the null hypothesis in favor of the alternative. Do not reject the null hypothesis in favor of the alternative.
Step 6. Interpretation:
At 1% significance level we Select an answer DO NOT DO have sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.
In: Statistics and Probability
Suppose that a foreign citizen and resident is considering immigration into the United States and owns both appreciated assets (value greater than basis) and depreciated assets (value less than basis) that she is thinking of selling. What practical tax planning steps should be suggested?
In: Accounting
For a certain drug, based on standards set by the United States Pharmacopeia (USP) - an official public standards-setting authority for all prescription and over-the-counter medicines and other health care products manufactured or sold in the United States, a standard deviation of capsule weights of less than 0.8 mg is acceptable. A sample of 20 capsules was taken and the weights are provided below:
| 120.3 | 120.8 | 120.1 | 119.7 | 120.8 |
| 119.4 | 119.1 | 120.9 | 118.9 | 119.5 |
| 120.4 | 121.1 | 118.6 | 119.4 | 119.3 |
| 119.8 | 120.2 | 119.5 | 118.9 | 119.8 |
(Note: The average and the standard deviation of the data are respectively 119.8 g and 0.73 g.)
At 5% significance level, test the claim that the standard deviation of capsule weights of the drug is different from 0.8 g.
Procedure: Select an answer One mean Z Hypothesis Test One mean T Hypothesis Test One proportion Z Hypothesis Test One variance χ² Hypothesis Test
Assumptions: (select everything that applies)
Step 1. Hypotheses Set-Up:
| H0:H0: Select an answer p σ² μ = | , where ? p μ σ is the Select an answer population proportion population standard deviation population mean and the units are ? lbs g mg kg |
| Ha:Ha: Select an answer μ p σ² ? > ≠ < | , and the test is Select an answer Right-Tail Left-Tail Two-Tail |
Step 2. The significance level α=α= %
Step 3. Compute the value of the test statistic: Select an answer z₀ f₀ χ²₀ t₀ = (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 | ? yes no |
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 DO NOT have sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.
In: Statistics and Probability
In 1898 the United States and Spain fought a war over the U.S. intervention in the Cuban War of Independence. At that time the U.S. military was concerned about the nutrition of its recruits. Many did not have a sufficient number of teeth to chew the food provided to soldiers. As a result, it was likely that they would be undernourished and unable to fulfill their duties as soldiers. The require-ments at that time specified that a recruit must have "at least four sound double teeth, one above and one below on each side of the mouth, and so opposed" so that they could chew food. Of the 58953 recruits who were under the age of 20, 63 were rejected for this reason. For the 43786 recruits who were 40 or over, 3829 were rejected.
(a) Find the proportion of rejects for each age group ( ±± 0.0001).
pˆ<20p^<20 =
pˆ40+p^40+ =
(b) Find a 99% confidence interval ( ±± 0.001) for the difference in the proportions.
A 99% confidence interval is from to
(c) Use a significance test to compare the proportions ( ±± 0.001)
pˆp^ =
zz =
PP -value =
Conclusion
We have evidence to conclude there was an age group difference in the rejection rate
OR
We have no evidence to conclude that there was an age group difference in the rejection rate
In: Statistics and Probability
The mean income per person in the United States is $45,000, and the distribution of incomes follows a normal distribution. A random sample of 17 residents of Wilmington, Delaware, had a mean of $55,000 with a standard deviation of $8,500. At the 0.100 level of significance, is that enough evidence to conclude that residents of Wilmington, Delaware, have more income than the national average? State the null hypothesis and the alternate hypothesis. State the decision rule for 0.100 significance level. (Round your answer to 3 decimal places.) Compute the value of the test statistic. (Round your answer to 2 decimal places.) Is there enough evidence to substantiate that residents of Wilmington, Delaware, have more income than the national average at the 0.100 significance level?
In: Statistics and Probability
1. In the 1970s, the United States federal government created a Department of Energy. This is a time when the OPEC (Organization of Petroleum Exporting Countries) cartel first became prominent. Identify how this action might have impacted the three major macroeconomic goals of our economy.
2. Suppose you live in a community of 100 people where everyone is able and seeks to work. If 80 people are over 16 years old and 72 of them are employed, what is the unemployment rate in this community?
3. What are the three major types of unemployment? What are their causes?
4. What is the business cycle? Explain the four phases of the business cycle.
5. Suppose a consumer buys 10 units of good X and 20 units of good Y every year. The following table lists the prices of goods X and Y in the years 2005-2007. Assume that these two goods constitute the typical market basket. Calculate the price indices for these years with 2005 as the base year. Comment on the inflation picture for these years.
|
Year |
Good X |
Good Y |
|
2005 |
$3 |
$6 |
|
2006 |
4 |
7 |
|
2007 |
4.5 |
7.5 |
In: Economics
According to the US Census Bureau, the Gini Coefficient in the United States was 0.397 in 1967, and 0.480 in 2014. The Gini coefficient is a measure of inequality that ranges from 0 to 1, where higher numbers indicate greater inequality. According to the World Bank, countries with Gini coefficients between 0.5 and 0.7 are characterized as highly unequal. Using the idea that Incentives Matter, analyze BOTH the pros of some income inequality and the cons of excessive income inequality.
Some income inequality can be good because of the " trickle down effect" and rewarding hard work and taking risks and investments.
The cons of excessive income inequality are unfair monopoloy, homelessness, and can easily lead to poverty.
In: Economics
Cabinaire Inc. is one of the largest manufacturers of office furniture in the United States. In Grand Rapids, Michigan, it assembles filing cabinets in an Assembly Department. Assume the following information for the Assembly Department: Direct labor per filing cabinet 20 minutes Supervisor salaries $135,000 per month Depreciation $16,000 per month Direct labor rate $18 per hour Prepare a flexible budget for 9,000, 11,000, and 14,000 filing cabinets for the month of August in the Assembly Department, similar to Exhibit 5. Assuming that inventories are not significant. Enter all amounts as positive numbers. CABINAIRE INC-ASSEMBLY DEPARTMENT Flexible Production Budget For the Month Ending August 31 (assumed data) Units of production 9,000 11,000 14,000 Variable cost: Direct labor $ $ $ Total variable cost $ $ $ Fixed cost: Supervisor salaries $ $ $ Depreciation Total fixed cost $ $ $ Total department cost
In: Accounting
It appears that over the past 50 years, the number of farms in the United States declined while the average size of farms increased. The following data provided by the U.S. Department of Agriculture show five-year interval data for U.S. farms. Use these data to develop the equation of a regression line to predict the average size of a farm by the number of farms Discuss the slope and y-intercept of the model. Year Number of Farms (millions) Average Size (acres) 1950 5.70 209 1955 4.63 262 1960 3.91 296 1965 3.35 336 1970 2.95 374 1975 2.51 421 1980 2.45 425 1985 2.32 441 1990 2.15 459 1995 2.07 469 2000 2.17 433 2005 2.11 444 2010 2.19 420
In: Statistics and Probability
The Statistical Abstract of the United States published by the U.S. Census Bureau reports that the average annual consumption of fresh fruit per person is 99.9 pounds. The standard deviation of fresh fruit consumption is about 30 pounds. Suppose a researcher took a random sample of 38 people and had them keep a record of the fresh fruit they ate for one year.
(Round all z values to 2 decimal places. Round your answers to 4 decimal places.)
a. What is the probability that the sample average would be less than 90 pounds?
p =
b. What is the probability that the sample average would be between 98 and 105 pounds?
p =
c. What is the probability that the sample average would be less than 112 pounds?
p =
In: Statistics and Probability