A small shop is trying to compare the proportion of consumers who prefer buying a pre-made bread and those who prefer fresh bread. They randomly sample daily consumers and asked for their opinion. The shop is then decided to construct a 90% confidence interval for the difference of population proportion and found it equal to (-0.53 , 0.30). i. based on the interval, what is the decision for testing H oP-P2=0 against H 1;p,- P,#0 at 5% significant levels Ocannot be determine OFail to reject the null hypothesis OReject the null hypothesis

A small shop is trying to compare the proportion of consumers who prefer buying a pre-made bread and those who prefer fresh bread. They randomly sample daily consumers and asked for their opinion. The shop is then decided to construct a 90% confidence interval for the difference of population proportion and found it equal to (-0.53 , 0.30). i. based on the interval, what is the decision for testing H oP-P2=0 against H 1;p,- P,#0 at 5% significant levels Ocannot be determine OFail to reject the null hypothesis OReject the null hypothesis

Name: A small shop is trying to compare theproportion of consumers who pref
Uploaded: 2024-03-20T07:07:00.000Z
Channel: Bartleby
Description: A small shop is trying to compare theproportion of consumers who prefer buying apre-made bread and those who prefer freshbread. They randomly sample daily consumersand asked for their opinion. The shop is thendecided to construct a 90% confidenceint

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

Suppose you take 4 air quality samples in parts per million: 19.6, 1.88, 12.7, 11.9. Give a 90% confidence interval of the mean air quality? (Also, t(0.95,3) = 2.3534 and t(0.90,4)= 1.5332) (a) (3.45, 16.83) (b) (0, 25) (c) (2.93, 20.10) (d) (5.22, 21.54) (e) None of the above
A researcher is interested in studying the use of hypnosis to relieve pain. She took a random sample of 16 subjects and measured their sensory ratings. The mean of the sensory ratings was found to be 7.5 and standard deviation to be 2.3. It is known that the sensory ratings are normally distributed. Construct a 99% confidence interval for the mean sensory ratings of the population. a) (5.81, 9.19) b) (6.27, 8.73) c) (6.36, 8.64) d) (6.02, 8.98)
Pedro was working on a report on Swiss and Austrian psychiatrists. He decided to find a 90 percent confidence interval for the difference in mean age at the time of significant psychiatric discoveries for Swiss versus Austrian psychiatrists. He found the ages at the time of psychological discovery of all the members of both groups and found the 90 percent confidence interval based on a t-distribution using a calculator. The procedure he used is not appropriate in this context because O The sample sizes for the two groups are not equal. O The entire population is measured in both cases, so the actual difference in means can be computed and a confidence interval should not be used. O Age at the time of psychological discovery occurs at different intervals in the two countries, so the distribution of ages cannot be the same. O Age at the time of psychological discovery is likely to be skewed rather than bell shaped, so the assumptions for using this confidence interval formula are not…
Are the results between the two confidence intervals very different?
A beverage company would like to determine if their drink is preferred to a rival company’s drink. They perform an experiment with 150 randomly selected individuals and have them taste test the two different drinks in a controlled environment so that each person's preference is independent of others. A 90% confidence interval for the proportion preferring their drink is (0.453, 0.587). Which of the following is the correct interpretation of this confidence interval? 90% of all consumers prefer this beverage company’s drink to their rival’s between 45.3% and 58.7% of the time. We’re 90% confident that between 45.3% and 58.7% of all consumers would prefer this beverage company’s drink to their rival’s drink. We’re 90% confident that there’s a 45.3% to 58.7% chance that a randomly selected person would prefer this beverage company’s drink to their rival’s drink. We’re 90% confident that the proportion of people in the sample who prefer this beverage company’s drink over their…
Echinacea is widely used as an herbal remedy for the common cold, but does it work? In a double-blind experiment, healthy volunteers agreed to be exposed to common-cold-causing rhinovirus type 39 and have their symptoms monitored. The volunteers were randomly assigned to take either a placebo or an echinacea supplement daily for 5 days following viral exposure. Among the 103 subjects taking a placebo, 88 developed a cold, whereas 44 of the 48 subjects taking echinacea 1 developed a cold.
2 different treatments were used to speed the relief time (in minutes) of migraines. 8 subjects received each treatment. The recovery times are listed below. Treatment 1: 20 19 29 25 34 32 39 38 Treatment 2: 18 18 16 17 29 28 21 37 Construct a 90% confidence interval for the difference in the relief time between the 2 treatments.
The manager of a fleet of automobiles is testing two brands of radial tires and assigns one tire of each brand at random to the two rear wheels of eight cars and runs the cars until the tires wear out. The data (in kilometers) follow. Find a 99% confidence interval on the difference in the mean life. Car Brand 1 Brand 2 1 36,925 34,318 2 45,300 42,280 3 36,256 35.548 4 32,100 31,950 5 37,210 38,015 6 48,360 47,800 7 38.200 37,810 8 33,500 33,215 Round your answer to 2 decimal places. Do not use commas.
Construct the 99% confidence interval for the difference m1-m2 when x bar = 476.88, x bar = 318.32, s1=42.11, s2= 26.80, n1=19, and n2=22. Use the ti 84 calculator and round the answers to two decimal places. A 99% confidence interval for the difference in the population means is ?
Two brands of tires were compared for the proportion of defective tires. Random samples of tires were collected from each brand, and the number of defective tires from each brand was recorded. The data is in the following table. Number of defective tires (x) 41 32 n Continental Michelin 320 180 a.Construct a 97% confidence interval for the difference between the two proportions. b.For part a, find the desired sample sizes if you want the Continental sample to be 1.5 times Michelin sample with a sampling error equal to 0.03. c. Conduct a test of hypothesis to determine if the defective rates are the same. Use the P-value to interpret the results (a is not given).
Construct both 95% and 90% confidence interval for B, based on the information given BA, = -8.4, SSE = 146, SSxx = 64, %3D n = 20 %3D
You want to find a 92% confidence interval for the proportion of PLM students who get a student scholarship allowance from the government of the City of Manila. You found an information from the city administrator that states that around 75% of PLM students are to get the allowance. At a maximum error of 5%, how many samples do you need?