5.58 The concentration of mercury in a lake has been monitored for a number of years.Measurements taken on a weekly basis yielded an average of 1.20 mg/m (milligrams percubic meter) with a standard deviation of .32 mg/m. Following an accident at a smelter on theshore of the lake, 15 measurements produced the following mercury concentrations.1.601.771.61 1.08 1.07 1.79 1.341.071.451.59 1.43 2.07 1.16 0.85 2.11ES ABOUT POPULATION CENTRAL VALUESa. Give a point estimate of the mean mercury concentration after the accident.b. Construct a 95% confidence interval on the mean mereury concentration afterthe accident. Interpret this interval.c. Is there sufficient evidence that the mean mercury concentration has increasedsince the accident? Use a = .05.d. Assuming that the standard deviation of the mercury concentration is .32 mg/m.calculate the power of the test to detect mercury concentrations of 1.28, 1.32.1.36, and 1.40. QUESTION 85.58d: What is the power of the test for Ha =1.40 ? Use 3 decimals for youranswerHint: L-13, p12-13

Here as per POLICY I HAVE CALCULATED 3 SUBPART Here given data of 15 mercury concentration after…

Answered: a. Give a point estimate of the mean…

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

Assume that population mean is to be estimated from the sample described. Use the sample results to approximate the margin of error and 95% confidence interval. n=49, x=55.1 seconds, s=4.8 seconds The margin of error is nothing seconds.
Determine each of the following areas and show these graphically. Use probability notation in your final answer.a. above z = 1.46b. below z = - 0.50c. between z = 2.20 and z =1.00
Find the critical t-value that corresponds to 90% confidence and n = 15. A. 1.761 OB. 2.624 O C. 1.345 O D. 2.145 18
Now change the 'Normal' choice to 'Exponential' This changes the underlying population from one that has a normal distribution to one that is very not normal. Change the sample size to 5 and run samples. a. How well do the 95% confidence intervals do at capturing the true population mean when samples sizes are small? b. Now change the sample size to 40 and run samples. Does a larger sample size mean that the intervals are more likely to capture the true population value? Why? Note THIS is an important concept and relates back to the Sampling Distribution of Sample Means and how the SDSM changes as sample size increases when the population is not normal.
You are a researcher studying a certain species of fish in a local lake. You gather a random sample of 19 fish of this species and record the lengths (in centimeters) of the fish. The data is displayed in the table below. Estimate the mean length of all fish of this type in the lake. Use a 99% confidence level. 9.1 9 8.6 5.8 9.8 7.4 9.1 9.5 8.2 8.9 9.3 7.8 9 9.6 10.4 13.6 7.2 10.1 8.9 a) Find the confidence interval: b) What is the margin of error of your interval? c) Interpret your 99% confidence interval in context
Consider a value to be significantly low if its z score less than or equal to −2 or consider a value to be significantly high if its z score is greater than or equal to 2. A test is used to assess readiness for college. In a recent year, the mean test score was 21.4 and the standard deviation was 5.4. Identify the test scores that are significantly low or significantly high. What test scores are significantly low? Select the correct answer below and fill in the answer box(es) to complete your choice. A. Test scores that are between ____ and _____. (Round to one decimal place as needed. Use ascending order.) B. Test scores that are less than _____. (Round to one decimal place as needed.) C. Test scores that are greater than _____. (Round to one decimal place as needed.)
Tina catches a 14-pound bass. She does not know the population mean or standard deviation. So she takes a sample of five friends and they say the last bass they caught was 9, 12, 13, 10, and 10 pounds. Find the t and calculate a 95% (α = .05) confidence interval.
Find the value:
A statistics class is estimating the mean height of all female students at their college. They collect a random sample of 42 female students and measure their heights. The mean of the sample is 65.3 inches and the sample standard deviation is 5.4 inches. Find the 90% confidence interval for the mean height of all female students in their school? Assume that the distribution of individual female heights at this school is approximately normal. 1. The t value is 1.3025 2. The standard error has a value of -- 3. The margin of error has a value of 4. Find the 90% confidence interval for the mean height of all female students in their school
Use z scores to compare the given values. The tallest living man at one time had a height of 238 cm. The shortest living man at that time had a height of 142.4 cm. Heights of men at that time had a mean of 175.45 cm and a standard deviation of 5.59 cm. Which of these two men had the height that was more extreme? ... Since the z score for the tallest man is z = 0 and the z score for the shortest man is z = the man had the height that was Im- more extreme. (Round to two decimal places.) shortest tallest
A quiz related to digital knowledge was conducted. The quiz had 10 questions and covered topics such as the purpose of browser cookies, phishing scams, and privacy policies. The survey was given to 50 people. The mean score is 4.1 with a standard deviation of 2.6. We want to know if the data provide evidence that the mean score in the population is lower than 5. We will use alpha = 0.05 to make our decision. part a. Is this a one-sided or two-sided test? part b. What is the standard error for this problem? part c. How many degrees of freedom do we have for this problem? part d. What is the test statistic, t, for this problem? Based on your test statistic and degrees of freedom, the p-value will be between what two values? (If you cannot find the exact df in the table, round down to be safe. So if you have 38 df, round down to 30df.) Group of answer choices The p-value is between .1 and .05 The p-value is between .05 and .025 The p-value is between .025 and .01 The p-value…
Consider a set of data in which the sample mean is 30.230.2 and the sample standard deviation is 7.87.8. Calculate the z-score given that x=31.4x=31.4. Round your answer to two decimal places.

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