**Question: Improve Understanding of Confidence Intervals and Margin of Error**A sample of $ n = 25 $ diners at a local restaurant had a mean lunch bill of $16 with a standard deviation of $ \sigma = \$4 $. We obtain a 95% confidence interval as (14.43, 17.57). Which action will not reduce the margin of error?- ○ Sampling from a population with less variability- ○ Decreasing the sample size- ○ Decreasing the confidence level- ○ Increasing the number of digits for calculating the interval**Explanation for Options:**1. **Sampling from a population with less variability:** This typically reduces the margin of error since less variability means more reliable estimates.2. **Decreasing the sample size:** Generally, this increases the margin of error because smaller samples tend to provide less precise estimates, so this is the correct answer.3. **Decreasing the confidence level:** This usually reduces the margin of error as a lower confidence level means a narrower interval.4. **Increasing the number of digits for calculating the interval:** This doesn't affect the actual statistical margin of error; it merely changes the precision of how the results are reported.**Tip:** When assessing actions in statistical analysis, consider how each factor influences the overall precision and reliability of your estimates.

Name: **Question: Improve Understanding of Confidence Intervals and Margin
Uploaded: 2024-03-20T07:07:12.000Z
Channel: Bartleby
Description: **Question: Improve Understanding of Confidence Intervals and Margin of Error**A sample of $ n = 25 $ diners at a local restaurant had a mean lunch bill of $16 with a standard deviation of $ \sigma = \$4 $. We obtain a 95% confidence interval as

Sample size , n = 25 Mean = 16 Standard deviation = 4 95% confidence interval as (14.43, 17.57).

Math

Statistics

A sample of n = 25 diners at a local restaurant had a mean lunch bill of $16 with a standard deviation of o = $4. We obtain a 95% confidence interval as (14.43, 17.57). Which action will not reduce the margin of error? O sampling from a population with less variability O decreasing the sample size decreasing the confidence level increasing the number of digits for calculating the interval

A sample of n = 25 diners at a local restaurant had a mean lunch bill of $16 with a standard deviation of o = $4. We obtain a 95% confidence interval as (14.43, 17.57). Which action will not reduce the margin of error? O sampling from a population with less variability O decreasing the sample size decreasing the confidence level increasing the number of digits for calculating the interval

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

Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 85 weekly reports showed a sample mean of 18.5 customer contacts per week. The sample standard deviation was 5.4. Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel. 90% confidence interval, to 2 decimals: 95% confidence interval, to 2 decimals:
Mrs. Gallas wanted to estimate the average weight of an Oreo cookie to determine if theaverage weight was less than advertised. She selected a random sample of 30 cookies andfound the weight of each cookie (in grams). The mean weight was = 11.1921 grams with astandard deviation of sx = 0.0817 grams. Make a 95% confidence interval to estimate the truemean weight of an Oreo. 1. What is the formula for calculating the standard deviation of the sampling distribution of x? 2.What condition must be met to use this formula? Has it been met?
Traumatic Brain Injuries have been shown to affect brain volume. A psychologist measures the brain volume of N=8 TBI patients at 2 months and 12 months after their injuries. Evaluate the results of her data using α = .052tail. **Attached Image** a. Show how the 95% confidence interval was calculated. b. State the conclusion in APA format.
Aisha wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 72 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 6.5 and a standard deviation of 2.7. What is the 80% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Assume the data is from a normally distributed population. Round answers to 3 decimal places where possible.
Work Time Lost due to Accidents At a large company, the Director of Research found that the average work time lost by employees due to accidents was 102 hours per year. She used a random sample of 21 employees. The standard deviation of the sample was 5.4 hours. Estimate the population mean for the number of hours lost due to accidents for the company, using a 99% confidence interval. Assume the variable is normally distributed. Use a graphing calculator and round the answers to the nearest whole number.
Isabella wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 62 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 7.8 and a standard deviation of 1.3. What is the 90% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Assume the data is from a normally distributed population. Round answers to 3 decimal places where possible.
Use the given confidence interval to find the margin of error and the sample mean. (13.2,20.6) The sample mean is The margin of error is (Type an integer or a decimal ) (Type an integer or a decimal)
Overweight Men For a random sample of 70 overweight men, the mean of the number of pounds that they were overweight was 29. The standard deviation of the population is 4.3 pounds. Find the 90% confidence interval of the mean of these pounds. Round intermediate answers to at least three decimal places.
The z confidence interval for the mean of a Normal population will have a specified margin of error m when the sample size is: a. n = (=*o/m). b. n = (z/m)². O c. n = (*o/m)². d. n ==*o/m.
Determine the margin of error, m, for a 95% confidence interval when the sample size is n = 19 and the sample standard deviation is s = 7. Give your answer precise to three decimal places. m =