3. A recent survey indicated that 78% of secondary-school students graduate witin 5 years of entering grade 9.The result is considered accurate to within a margin of error of ±4%, 19 times out of 20.(a)Find the confidence interval width, w.(b)What value for za/2 is suggested by the phrase "19 times out of 20"?(c)Identify p and q.(d)Estimate the sample size n that this survey must have used.

Answered: Image /qna-images/answer/a599210c-550a-420d-99e6-311f3f74e100.jpg

Answered: 3. A recent survey indicated that 78%…

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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A random sample of 20 grade 1 students in Pio Elementary School yielded a mean weight of 31.68 kg with sample standard deviation of 6.39 kg. What is the maximum error of 95% confidence interval for the mean weight of all grade 1 students in Pio Elementary School. Round off your answer to two decimal places
#21)
A normally distributed population is having standard deviation 2.4, a sample of 24 is taken randomly and the sample mean observed is 13.8 Find the 95% confidence interval. (round to 3 decimals)
A simple random sample of size n=24 is drawn from a population that is normal distributed. The sample mean is found to be x bar= 66 and the sample standard deviation is found to be s= 16.construct a 98% confidence interval about the population mean. (Round to two decimals)
Below is a 75% confidence interval for a population proportion from a sample of size 90 with sample proportion 59 90 ( 0.598 , 0.713 ) Determine the margin of error for this confidence interval. margin of error = [three decimal accuracy]
What price do farmers get for their watermelon crops? In the third week of July, a random sample of 40 farming regions gave a sample mean of x = Assume that o is known to be $1.92 per 100 pounds. $6.88 per 100 pounds of watermelon. (a) Find a 90% confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop (in dollars). What is the margin of error (in dollars)? (For each answer, enter a number. Round your answers to two decimal places.) lower limit $ upper limit 2$ margin of error $ (b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.27 for the mean price per 100 pounds of watermelon. (Enter a number. Round up to the nearest whole number.) farming regions (c) A farm brings 15 tons of watermelon to market. Find a 90% confidence interval for the population mean cash value of this crop (in dollars). What is the margin of error (in dollars)? Hint: 1 ton is 2000 pounds. (For each…
Prof. Gersch knows that the score on a standardized test is perfectly normal with unknown mean and a standard deviation of 50. He then takes a random sample of 25 tests and finds that the average of these is 200. How many exams should he sample so that the maximum margin of errorfor his 95% confidence interval is 10?: a. 98 b. 97 c. 100 d. 96 e. 10
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.
You want to estimate the mean weight of quarters in circulation. A sample of 40 quarters has a mean weight of 5.629 grams and a standard deviation of 0.063 gram. Use a single value to estimate the mean weight of all quarters. Also, find the95% confidence interval for the average weight of all quarters. The estimate for the mean weight of all quarters is 5.6295.629 grams. (Round to three decimal places as needed.) Find the 95% confidence interval for the mean weight of all quarters. nothing grams<μ<nothing grams
A sample of 14 male college students consumes an average of 14.3 gallons of beer per month, with a sample standard deviation of 5.2 gallons. You can safely assume that the population of monthly beer consumption is Normally distributed. What is the margin of error for a 90% Confidence Interval for the mean? (Leave your answer to two decimal places, ex: 18.12)
The data below were collected in a clinical trial to compare a new drug to a placebo for its effectiveness in lowering total serum cholesterol. Generate a 95% confidence interval for the difference in mean total cholesterol levels between treatments. Generate a 95% confidence interval for the proportion of all patients with total cholesterol < 200. New Drug Placebo Total Sample (n=75) (n=75) (n=150) Mean (SD) Total Serum Cholesterol 185.0 (24.5) 204.3 (21.8) 194.7 (23.2) % Patients with Total Cholesterol < 200 78.00% 65.00% 71.50% a. 95% CI ( , )
A sample of size n=79 Is drawn from a population whose standard deviation is o = 29, Part 1 of 2 (a) Find the margin of error for a 90% confidence interval for l. Round the answer to at least three decimal places. The margin of error for a 90% confidence interval for u is Part 2 of 2 (b) If the sample size weren=44, would the margin of error be larger or smaller? (Choose one) ▼ because the sample size is (Choose one) Larger Smaller

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