Week 5 Assignment
In a poll of 725 voters in a campaign to eliminate non-returnable beverage containers, 510 of the voters were opposed. Develop a 95% confidence interval estimate for the proportion of all the voters who opposed the container control bill.
A random sample of 25 airline pilots had an average yearly income of $101,600 with a sample standard deviation of $10,000.
- If we want to determine a 99% confidence interval for the average yearly income, what is the value of t?
- What wording in the prompt tells us to use t instead of z?
- Develop a 99% confidence interval for the average yearly income of all pilots.
In order to determine the average weight of carry-on luggage by passengers in airplanes, a sample of 425 pieces of carry-on luggage was collected and weighed. The average weight was 17 pounds. Assume that we know the standard deviation of the population to be 2.75 pounds.
- Determine a 90% confidence interval estimate for the mean weight of the carry-on luggage.
- What wording in the prompt gave away the type of test (z or t) that this should be?
A statistician employed by a consumer testing organization reports that at 90% confidence he has determined that the true average content of the Uncola soft drinks is between 11.8 to 12.2 ounces. He further reports that his sample revealed an average content of 12 ounces, but he forgot to report the size of the sample he had selected. Assuming the standard deviation of the population is 0.72, determine the size of the sample.
Hint from Dr. Klotz: Have you visited this week’s Notations and Symbols to align the formulas with these problems?
|Grading Criteria Assignments||Maximum Points|
|Meets or exceeds established assignment criteria||40|
|Demonstrates an understanding of lesson concepts||20|
|Clearly presents well-reasoned ideas and concepts||30|
|Uses proper mechanics, punctuation, sentence structure, and spelling||10|