CI examples_Review.pdf - Compare two problems below what...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Compare two problems below: what parameter we intend to estimate? I. Sample of 400 randomly selected flight reservations was chosen. 10 of them were canceled on the day of the flight. Find a 98% confidence interval for the true percentage of flight reservations being canceled on the day of the flight. Hint: Qualitative Data Parameter: P II. Sample of 50 days was randomly chosen and percentage of canceled flight reservations was recorded for each day. Find a 94% confidence interval for the average percentage of flight reservations being canceled on the day of the flight. Hint: Quantitative Data Parameter: µ Confidence Interval for a Population Mean 1. A physician wanted to estimate the mean length of time μ that a patient had to wait to see him after arriving at the office. A random sample of 36 patients showed a mean waiting time of 23.4 minutes and a standard deviation of 7.2 minutes. Find a 96% confidence interval for μ . Parameter of interest is Population Mean ( μ ) Point – estimator is sample mean : min 23.4 = x To find critical value 2 α Z take half of confidence level (.96:2 = .48), then use Normal table to locate entry closest to 0.4800----> 0.4798, which corresponds to Z-score = 2.05 CI: n S Z X 2 α ± = 23.4 ± 36 2 . 7 05 . 2 = 23.40 ± 2.46 SE = 2.46 Conclusion: We are 96% confident that average time to wait in physician office is within (22.94, 25.86) min 2. The quality control officer collects a random sample of 49 yardsticks from the day's production run. The sample mean is 36.00 inches and the standard deviation is 1.4 inches. Find a 99% confidence interval for the mean length of all yardsticks made that day. Parameter of interest is Population Mean ( μ ) Point – estimator is sample mean : in 36 = x To find critical value 2 α Z take half of confidence level (.99:2 = .4950), then use Normal table to locate entry closest to 0.4950---> 0.4945 and 0.4955, compute average between Z = 2.57 and Z = 2.58 which corresponds to Z-score = 2.575 CI: n S Z X 2 α ± = 36.0 ± 49 4 . 1 575 . 2 = 36 ± .52 SE = .52 Conclusion: We are 99% confident that average length of all yardsticks made that day is within (35.48, 36.52) in.
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
3. During the last Super-Bowl Sunday, Adrian and his buddies ordered 27 pizzas from Pizzas To Go at different times. The average delivery time proved to be 23.7 min, with a standard deviation of 10.7 min. Assume a normal distribution. Feeling this was far too long a delay, Adrian and his friends decided to buy the 28th pizza elsewhere if it appeared that the delivery time for Pizzas To Go exceeded 30 min. Set alpha at 1%. Will they order elsewhere? Hypothesized Parameter value (Population Mean): µ o = 30 min Point –estimator (sample mean): min 23.7 = x CI: n S t X 2 α ± = 23.7 ± 27 7 . 10 779 . 2 = 23.7 ± 5.72 Answer: (17.98, 29.42) min. Interpretation: You can be 99% confident that the pizzas from Pizzas To Go will arrive in less than 30 minutes, since 30 does not fall in the interval. Continue ordering from Pizzas To Go.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern