In a poll of 500 voters in a campaign to eliminate non-returnable beverage containers, 275 of the voters were opposed. Develop a 95% confidence interval estimate for the proportion of all the voters who opposed the container control bill.
Need assignment help for this question?
If you need assistance with writing your essay, we are ready to help you!
Why Choose Us: Cost-efficiency, Plagiarism free, Money Back Guarantee, On-time Delivery, Total Сonfidentiality, 24/7 Support, 100% originality
A random sample of 87 airline pilots had an average yearly income of $97,000 with a standard deviation of $3,000.
- If we want to determine a 95% confidence interval for the average yearly income, what is the value of t?
- What are the degrees of freedom for this problem, and how is this value calculated?
- Develop a 95% 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 100 pieces of carry-on luggage was collected and weighed. The average weight was 34 pounds. Assume that we know the standard deviation of the population to be 5 pounds.
- Determine a 90% confidence interval estimate for the mean weight of the carry-on luggage.
- What determined whether you used a t value or a z value?
A statistician employed by a consumer testing organization reports that at 95% confidence he has determined that the true average content of the Uncola soft drinks is between 11.9 to 12.1 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 0.5, determine the size of the sample.