**Example: Calculating the Margin of Error****Problem:** Assume that a sample is used to estimate a population mean \( \mu \). Find the margin of error \( M.E. \) that corresponds to a sample of size 22 with a mean of 49.8 and a standard deviation of 5.6 at a confidence level of 90%.**Instructions:** Report \( M.E. \) accurate to one decimal place because the sample statistics are presented with this accuracy.**Calculation:** \[ M.E. = \] [Input Box]**Note:** The answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.For educational purposes, the formula for \( M.E. \) using the appropriate critical value for a 90% confidence interval should be used:\[ M.E. = t^* \times \left(\frac{s}{\sqrt{n}}\right) \]where:- \( t^* \) is the critical value from the t-distribution table- \( s \) is the sample standard deviation- \( n \) is the sample size

Answered: Assume that a sample is used to…

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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Similar questions

The mean score on a driving exam for a group of driver's education students is 6868 points, with a standard deviation of 55 points. Apply Chebychev's Theorem to the data using kequals=22. Interpret the results
The average height of students at a university from an SRS of 14 students gave a standard deviation of 2.5 feet. Construct a 95% confidence interval for the standard deviation of the height of students at the university. Assume normality for the data.
A veterinarian wants to estimate the mean weight of a cat in their city. They take a random sample of 19 cats and found a mean of 10.9 lbs with a sample standard deviation of 1 lbs. Find the 95% confidence interval of the mean. Assume that the cat weights are normally distributed. Do not round in between steps. Round answers to at least 4 decimal places.
The heights of 9 men were taken with sample mean of 165 cm and standard deviation 5 cm. The heights of 7 Japanese women were taken with sample mean of 150 cm and standard deviation 3 cm. Assuming that the population standard deviations for heights of men and women are both the same, construct the 95% confidence interval for the difference in mean heights between men and women.
In Arkansas, bass fish were collected in 30 different lakes to measure the amount of mercury in the fish. The mean amount of mercury was = 0.6443 with standard deviation s = 0.3606. Estimate the mean amount of mercury in fish in Arkansas lakes using a 95% confidence level. Round your answers to three decimal places, and use four decimal places for any interim calculations. With 95 % confidence, it can said that the true mean amount of mercury in fish in Arkansas lakes is between 0.502 x and 0.786 X Hint: Make sure you round your answer to three decimal places.
The mean score in miss smiths 4th grade class was 76, with a standard deviation of 7. Assume the grades are normally distributed and use the empirical rule to find the approximate percentage of students who scored above 83.
The sample mean of depression scores of 16 fifth graders was 4.4 with standard deviation of 8. If the population mean of this depression score is 6, what is the calculated t score for this sample?
The scores on a standardazed test are normally distributed with a mean of 115 and standard deviation of 7. What test score is 0.45 standard deviations below the mean?
The Weather Underground reported that the mean amount of summer rainfall for the northeastern US is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the α = 0.05 level, can it be concluded that the mean rainfall was below the reported average? What if α = 0.01? Assume the amount of summer rainfall follows a normal distribution. Please also state if the assumption of normal distribution was essential for solving this question. You will need to specify the appropriate hypothesis, identify the statistical test, estimate p-value (using R command), and state your conclusion in the context of the problem. For the conclusion, please use the following template. If you reject the null writeSince p-value is [write the value of your p-value] and is less than [write the value of alpha inpercent] we reject the null hypothesis at [write the value of alpha]% significance…
The ponderal index is a measure of overall size similar to a body mass index. The ponderal index of full term newborns follows a normal distribution with a mean of 2.4 kg/m3 and a standard deviation of 0.45 kg/m³. What percentage of full term newborns have a ponderal index greater than 2.5 kg/m³? Use Table 3 to obtain your answer. If necessary, round your final answer to 2 places after the decimal. Do not enter the % symbol, just the number representing the percent.
In Idaho, bass fish were collected in 30 different lakes to measure the amount of mercury in the fish. The mean amount of mercury was x = 0.594 with standard deviation s = : 0.4329. Estimate the mean amount of mercury in fish in Idaho lakes using a 99 % confidence level. Round your answers to three decimal places, and use four decimal places for any interim calculations. With between % confidence, it can said that the true mean amount of mercury in fish in Idaho lakes is and

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