CI examples_Review.pdf

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

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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.

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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.
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