According to Keynote Systems, Wal-Mart's website was available 95% of the time during themost recent holiday shopping season. Assuming this finding to be based on a random sample of200 attempts, the 85% confidence interval for the population proportion of the time the Wal-Mart site was available during this period is0.9198 to 0.98020.9141 to 0.98590.9278 to 0.97220.9303 to 0.96970.9246 to 0.97544%5门>6A&7W* 00P8&cricutO

Confidence interval:The confidence interval (CI) is a range of values that’s likely to include a…

Answered: According to Keynote Systems,…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Similar questions

If n = 15, ¯xx¯ = 36, and s = 12, construct a confidence interval at a 90% confidence level. Assume the data came from a normally distributed population. Give your answers to three decimal places.
A survey was conducted to determine what percentage of college seniors would have chosen to attend a differentcollege if they had known then what they know now. In a random sample of 100 seniors, 34 percent indicated thatthey would have attended a different college. A 90 percent confidence interval for the percentage of all seniors whowould have attended a different college is
Construct a 95% confidence interval for p1 - p2 for a survey that finds 30% of 240 males and 41% of 200 females are opposed to the death penalty. Group of answer choices a.(-0.200, -0.021) b.(-1.532, 1.342) c.(-1.324, 1.512) d.(-0.561, 0.651)
in a town in nj we take a random sample of 100 residents. It turns out that 25% of the adult residents have degrees higher than high school diploma. Find a 90% confidence interval.
In a survey funded by the UW school of medicine, 750 of 1000 adult Seattle residents said they did not believe they could come down with a sexually transmitted infection (STI). Construct a 95% confidence interval estimage of the proportion of adult Seattle residents who don't believe they can contract an STI. (Use a z score of 1.96 for your computations.) (.728, .772) (.723, .777) (.718, .782) (.713, .878) (.665, .835)
In a random sample of 500 people eating lunch at a cafeteria on various Fridays, it was found that 160 preferred seafood. A 99% confidence interval for the actual proportion of people eating seafood on Fridays (rounded off to two decimals) is: A. [0.28; 0.36] B. [0.29; 0.35] C. [0.27; 0.37] D. [0.26; 0.36]
A recent survey of 1,000 people found that 90% of Americans support background checks for purchasing guns. Construct a 95 % confidence interval for the population proportion.
When Mendel conducted his famous genetics experiments with peas, one sample of offspring consisted of 428 green peas and 152 yellow peas. 3. a. Find a 99% confidence interval estimate of the percentage of yellow peas.
An insurance company collects data on seat-belt use among drivers in a country. Of 1500 drivers 30-39 years old, 33% said that they buckle up, whereas 458 of 1800 drivers 55-64 years old said that they did. Find a 95% confidence interval for the difference between the proportions of seat-belt users for drivers in the age groups 30-39 years and 55-64 years. ..... Construct a 95% confidence interval. The 95% confidence interval for p, - p2 is from to (Round to three decimal places as needed.)
In a sample of 100 college students, 25 of them own a car. Calculate the 95% confidence interval for the true population of college students who own a car. Choose the closest one. A. (0.165132, 0.334868) B. (0.231725, 0.452302) C. (0.25, 0.75) D. (0.1, 0.9)
Listed below are the amounts of net worth (in millions of dollars) of the ten wealthiest celebrities in a country. Construct a 95% confidence interval. What does the result tell us about the population of all celebrities? Do the data appear to be from a normally distributed population as required? 245 210 186 168 166 163 156 156 156 156
The average amount of money spent by an Amazon customer per month is $1296 with standard deviation $110. Assume the variable is normally distributed. Answer the following questions. What is the probability that a random sample of 15 Amazon customers spent an average between $1100 and $1300 in a month?

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