![treestump090](/avatars/4346.jpg)
treestump090
04.04.2020 •
Mathematics
A survey of 1000 air travelers1 found that prefer a window seat. The sample size is large enough to use the normal distribution, and a bootstrap distribution shows that the standard error is . Use a normal distribution to find a 90% confidence interval for the proportion of air travelers who prefer a window seat. Round your answers to three decimal places.
Solved
Show answers
More tips
- F Food and Cooking How to Cook Julienne? Recipes and Tips...
- D Dating, Love, Relationships 10 Useful Tips on How to Survive a Breakup?...
- F Food and Cooking Apple Cider Vinegar: The Ultimate Health and Beauty Solution...
- C Computers and Internet Е-head: How it Simplifies Life for Users?...
- F Family and Home How to Choose the Best Diapers for Your Baby?...
- F Family and Home Parquet or laminate, which is better?...
- L Leisure and Entertainment How to Properly Wind Fishing Line onto a Reel?...
- L Leisure and Entertainment How to Make a Paper Boat in Simple Steps...
- T Travel and tourism Maldives Adventures: What is the Best Season to Visit the Luxurious Beaches?...
- H Health and Medicine Kinesiology: What is it and How Does it Work?...
Answers on questions: Mathematics
- M Mathematics Solve for x: log(216)=(x-4)log(6) step by step please and thank you!...
- M Mathematics Twelve added to a number equals 18 find the number...
- M Mathematics Rewrite the expression using rational exponents...
- B Biology Abald eagle and a black bear both have four limbs with digits because they are both tetrapods, descendants of a four-limbed ancestor. in this comparison, the limbs of the eagle and...
- H History Choose the group who protested the Stamp Act. First Continental Congress British government an unofficial group of colonists Second Continental Congress...
- M Mathematics How do you determine what the x and y variables represent and write an equation to represent this situation...
Ответ:
90% confidence interval for the true proportion of air travelers who prefer the window seat is (0.575, 0.625)
Step-by-step explanation:
We have the following data:
Sample size = n = 1000
Proportion of travelers who prefer window seat = p = 60%
Standard Error = SE = 0.015
We need to construct a 90% confidence interval for the proportion of travelers who prefer window seat. Therefore, we will use One-sample z test about population proportion for constructing the confidence interval. The formula to calculate the confidence interval is:
Since, standard error is calculated as:
Re-writing the formula of confidence interval:
Here,
is the critical value for 90% confidence interval. From the z-table this value comes out to be 1.645.
Substituting all the values in the formula gives us:
Therefore, the 90% confidence interval for the true proportion of air travelers who prefer the window seat is (0.575, 0.625)
Ответ:
I failed first grade
Step-by-step explanation: