Convert answers to percents and round to 2 decimal places. Example: 0.387582 is 38.76%23. A simple random sample of 1825 people living in a small city was asked if they support anadditional local sales tax to help fund the local community college. Of these people, 1304 said thatthey support this tax. Find the sample statistic and based on a 95% confidence interval find themargin of error and confidence interval for all residents in this city who support the tax.confidenceintervalm.o.e. == d

Name: Convert answers to percents and round to 2 decimal places. Example: 0
Uploaded: 2024-03-20T07:07:12.000Z
Channel: Bartleby
Description: Convert answers to percents and round to 2 decimal places. Example: 0.387582 is 38.76%23. A simple random sample of 1825 people living in a small city was asked if they support anadditional local sales tax to help fund the local community college. O

Given Information: Sample size n=1825 Out of which x=1304 people said that they support the tax.…

Math

Statistics

Convert answers to percents and round to 2 decimal places. Example: 0.387582 is 38.76% 23. A simple random sample of 1825 people living in a small city was asked if they support an additional local sales tax to help fund the local community college. Of these people, 1304 sa they support this tax. Find the sample statistic and based on a 95% confidence interval find margin of error and confidence interval for all residents in this city who support the tax. confidence m.o.e. = interval = d

Convert answers to percents and round to 2 decimal places. Example: 0.387582 is 38.76% 23. A simple random sample of 1825 people living in a small city was asked if they support an additional local sales tax to help fund the local community college. Of these people, 1304 sa they support this tax. Find the sample statistic and based on a 95% confidence interval find margin of error and confidence interval for all residents in this city who support the tax. confidence m.o.e. = interval = d

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Use the study below for Part A, Part B, and Part C. A soft drink company surveys 500 shoppers and finds that 61% of shoppers prefer diet soda to regular soda. Part A: With a confidence level of 90%, what is the margin of error? Show your work and round to the nearest hundredth.Part B: If the sample size would increase to 1,000 respondents while keeping the rest of the variables the same, what would happen to the margin of error?Part C: Support your answer to Part B by finding the margin of error if 1,000 shoppers were surveyed while keeping the rest of the variables the same. Show your
An economist predicts that in the next 50 years gas will be 0.02 based on the last 7 months of 2015. What type of error did she make ? Extrapolation? Simpson’s paradox? Influential outlier? Volunteer sample?
In a random sample of 500 teenagers 12 to 17 years old, it was found that 330 have regular access to computers and the Internet. Find a point estimate of the true proportion of teenagers 12 to 17 years old who have regular access to computers and the Internet.
A magazine provided results from a poll of 2000 adults who were asked to identify their favorite pie. Among the 2000 respondents, 11% chose chocolate pie, and the margin of error was given as ±3 percentage points. Describe what is meant by the statement that "the margin of error was given as ±3 percentage points." Choose the correct answer below. A. The statement indicates that the study is 100% - 3% = 97% confident that the true population percentage of people that prefer chocolate pie is 11%. B. The statement indicates that the interval 11% ± 3% is likely to contain the true population percentage of people that prefer chocolate pie. C. The statement indicates that the study is only 3% confident that the true population percentage of people that prefer chocolate pie is exactly 11%. D. The statement indicates that the true population percentage of people that prefer chocolate pie is in the interval 11% ± 3%.
9. The results from a sample of 175 JAC students found the average number of YouTube videos watched per day was 6.2. A previous census showed that the standard deviation was 1.74. l. Calculate the 98% confidence interval to estimate the average number of YouTube videos watched per day for the entire JAC population. b. What would happen to the interval range if the level of confidence was increased from 98% to 99.9%? NOTE: you do not need to do the calculation (you just need to explain what happens).
Please see below for the question details. I got 34.32 for the interest, but I also got 19.68 and 32 when I reworked the problem; Please help!!!! Because of the relatively high interest rates, most consumers attempt to pay off their credit card bills promptly. However, this is not always possible. An analysis of the amount of interest paid monthly by a bank's Visa cardholders reveals that the amount is normally distributed with a mean of 27 dollars and a standard deviation of 8 dollars.Please use z-scores with at least 3 decimal places. D. What interest payment is exceeded by only 18% of the bank's Visa cardholders? Interest Payment =
Please answer the following parts d, e and f only, as a, b and c were alreday provided. Also, please show using graphing calculator and not Excel. Thanks. Adult IQ scores can be modeled using a normal model with mean 100 and a standadrd deviation of 15. a. Find the proportion of IQ scores that are less than 115. b. Find the proportion of IQ scores that are greater than 131.5. c. Find the proportion of IQ scores between 90 and 110. d. Find the proportion of IQ scores that are betewwn 110 and 120. e. Find the IQ score above which there are 10% of the scores. f. Find the IQ score below which there are 20% of the scores.
.
An auditor, examining a total of 820 accounts receivable of a corporation, took a random sample of 90 of them. The sample mean was $126.99, and the sample standard deviation was $44.21. Complete parts a through e. a. Using an unbiased estimation procedure, find an estimate of the population mean. An estimate of the population mean is $126.99. (Round to the nearest cent as needed.) b. Using an unbiased estimation procedure, find an estimate of the variance of the sample mean. An estimate of the variance of the sample mean is 19.357. (Round to three decimal places as needed.) c. Find a 90% confidence interval for the population mean. $119.68 <<$ 134.30 (Round to the nearest cent as needed.) d. A statistician found, for the population mean, a confidence interval running from $116.99 to $136.99. What is the probability content of this interval? The probability content is (Round to four decimal places as needed.)
Question 9 How could the survey question "Have you ever had a drink of alcohol?" be revised to obtain a valid inference about the percent of drinkers in the population? o Do your parents drink? O Do you like the taste of wine? How often do you have an alcoholic beverage? O What do you think of alcohol use? Question 10
Consider the following data: −10, −10, 0, 3, −10, 3, 3 Step 1 of 3: Calculate the value of the sample Variance. Round your answer to one decimal place. Step 2 of 3: Calculate the value of the sample deviation. Round your answer to one decimal place. Step 3 of 3: Calculate the value of the range.
A researcher has collected the following sample data. The mean of the sample is 5. 3 5 12 3 2 The 75th percentile is 5 8.5 9 10