was $2.62 per gallon.The following data show the price per gallon of regular gasoline for 15 randomly selected stations in the state.Complete parts a through c below.2.612.532.512.692.562.632.552.672.742.682.672.622.522.752.72a. Construct a 95% confidence interval to estimate the average price per gallon of gasoline in the state.The 95% confidence interval to estimate the average price per gallon of gasoline in the state is from $$ .to(Round to the nearest cent.)

Answered: Image /qna-images/answer/b071fea5-9aa6-4f97-8a4d-bdaeee560ea2.jpg

Answered: was $2.62 per gallon. The following…

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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Pulse Rates (beats per minute) Males Females 84 71 78 84 74 61 93 82 47 73 57 72 59 51 71 68 71 55 55 85 62 69 80 92 50 58 79 86 73 80 89 92 54 69 86 88 64 63 55 91 72 62 36 70 61 96 65 89 65 56 82 82 78 63 75 80 81 55 78 72 64 56 64 58 67 67 68 103 97 68 77 71 42 87 57 74 87 61 64 75
Assume that the population has a normal distribution. use a 95% Confidence level. In a survey of 3005 adults aged 57 through 85 years, it was found that 82% of them use at least one prescription medication. 1. How many people use at least one prescription medication. 2. Find the margin of error E that corresponds to a 95% confidence level. 3. Find the 95% confidence interval estimate of the population proportion p. 4. How many people must be surveyed if we want a confidence level of 99% and a margin of error of four percentage points? Use the sample proportion is 82%.
Please help me solve: A. the population portion B. the Margin of Error C. the confidence interval
find the sample size needed to estimate the percentage of adults who can wiggle their ears. use a margin of error of 4 percentage and use a confidence level of 95%. complete parts (a) and (b) below. b. assume that 21% of adults can wiggle their ears. b. Assume that 21% of adults can wiggle their ears
(in hours): 7, 6, 7, 8.5, 8, 9, 6.5 a. Find the mean, median, mode and standard deviation for their hours of sleep. b. Create a 99% confidence interval for the mean amount of hours they slept.
If [0.64, 0.82] is a confidence interval for the population proportion. Find the value of the point estimator of the population proportion. Select one: O a. 0.73 O b. 0.70 O c. 0.74 O d. 0.72
If [1740, 1850 is a confidence interval for a population mean. Find the margin of error. Select one: a. 60 O b. 80 C. 55 O d. 40
Directions: 1. Answer the following items. 2. Find the appropriate sample size given the confidence level, standard deviation, and margin of error. 3. Round off to the nearest whole number. 4. Just write the number, no other symbols or variables, no spacing or comma. Problem: E = 0.36 kg, s = 20 kg, 95% confidence level
Please answer all. Thank you. The width of a confidence interval for a proportion will be A.) wider for sample of size 100 than for a sample size of 50 B.) wider when the sample proportion is 0.50 than when the sample proportion is 0.20 C.) wider for 90 percent confidence than for 95 percent D.) narrower for 99 percent confidence than for 95 percent Which of the following is not a property of a normally distributed data. A.) the mean, median and mode coincide at the center B.) area under curve is 1 C.) the width of curve is determined by the mean of distribution. D.) the curve is symmetrical about its center.
Several studies showed that after eating a low-fat cereal for two meals a day, subjects had lost some weight.A cereal companyfinanced this research. Identify what is wrong. Choose the correct answer below. A. Since the research is composed of voluntary response samples, there may be key data points missing. B. The data used in the studies is not reliable because it was not measured by the administrator. C. It is questionable that the sponsor is a cereal company because this sponsor can be greatly affected by the conclusion. D. It is not possible to take accurate measurements.

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