(iv) The mean level of Mn in fruit juice samples (n = 5) detected by Mn absorbanceat 279.5 nm was found to be 20.934 ± 2.362 ng / g. Using this information reportthe experimental uncertainty based on a confidence interval of 50% and 95%.Comment on the effect of increased sampling on the confidence interval forvalues of the same mean and standard deviation.Degrees of Freedom 5011.00020.81630.7650.7410.72745Confidence level (%)959812.706 31.8214.3036.9653.1824.5412.7763.7472.5713.365906.3142.9202.3532.1322.0159963.6569.9255.8414.6044.03299.5127.32114.0897.4535.5984.773100636.57831.59812.9248.6106.869

Given Information: The mean level of Mn in fruit juice detected by Mn absorbance at 279.5 nm was…

Answered: (iv) The mean level of Mn in fruit…

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 population with μ of values has a normal distribution 10.8 and 97.6. You intend to draw = = o = u a random sample of size n = 122. What is the mean of the distribution of sample means? Hã = What is the standard deviation of the distribution of sample means? (Report answer accurate to 2 decimal places.) 07 = x
#4 Are women and men equally absent at PS 160? Test at .10 significance level. Women: Average number of days absent = 14.7; standard deviation = 5.0; n = 12 Men: Average number of days absent = 8.3; standard deviation = 4.0; n = 18
Two samples are taken with the following sample means, sizes, and standard deviationsx¯1 = 27 x¯2 = 29n1 = 65 n2 = 72s1 = 4 s2 = 5Estimate the difference in population means using a 87% confidence level. Use a calculator, and do NOT pool the sample variances. Round answers to the nearest hundredth.______< μ1−μ2_____
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You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately 80.3. You would like to be 95% confident that your estimate is within 1 of the true population mean. How large of a sample size is required? Use a critical value rounded to three decimal places.
The viscosity of a liquid detergent is supposed to average 800 centistokes at 25°C. A random sample of 16 batches of detergent is collected, and the average viscosity is 812. Suppose we know that the standard deviation of viscosity is σ = 25 centistokes. (a) What is the P-value for the test? (b) Find a 95 percent confidence interval on the mean.
You want to obtain a sample to estimate a population mean.you believe the population standard deviations is approximately sigmma=23.3 you would like to be 95% confidence that your estimate is within 1.8 of the true population mean . How large of sample size is required ? The z critical value is given by 1.960
You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately σ=42.2. You would like to be 95% confident that your esimate is within 2 of the true population mean. How large of a sample size is required?n = Do not round mid-calculation. However, use a critical value accurate to six decimal places
Question 2 > You wish to test the following claim (Ha) at a significance level of a = 0.001. H.:µ = 50.6 H.:µ + 50.6 You believe the population is normally distributed and you know the standard deviation is o = 18.1 .You obtain a sample mean of M = 56.7 for a sample of size n = 46. What is the test statistic for this sample? (Report answer accurate to three decimal places.) 个 test statistic = What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = The p-value is.. O less than (or equal to) a O greater than a tv 80 88 DII DD esc F1 F2 F3 F4 F5 F6 F7 F8 F9 23
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A random sample of size 100 is taken from a normal distribution population with sigmma=4.3. give that the sample mean is x bar =19.8 construct a 95%confidence interval for the position mean mu.( Round to two decimals)
(b): Your answer is wrong. Assume that the population lengths of adult male killer whales follow approximately a normal distribution. A recent article published in the journal Zoology Now states that the mean of this population is6.99 m. You want to test the claim in the article, so you select a random sample of 10adult male killer whales and record the lengths of each. Follow these steps to construct a confidence interval for the90%for the population mean of all lengths of adult male killer whales. Then, indicate whether the confidence interval that you construct contradicts the article's statement. (If necessary, you can refer to a list of formulas .) (to) Click "Take Sample" to see the results of your random sample. Take a sample Sample size: 10 number of killer whales Point estimate: 7241 10 sample mean 7,241 sample standard Enter the values for the sample size, point estimate of the mean, sample standard deviation, and critical value that you need to construct the confidence…

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