A clinic offers a weight-loss program. The table below gives the amounts of weight loss, in pounds, for a random sample of 20 of its clients at the conclusion of the program. Assume that the data arenormally distributed. Complete parts (a) and (b)19 25 15 30 10 81491213 18 225131321 1119 1520 O30a Find a 90% confidence interval for the population meanThe 99% confidence interval is from a lower limt of to an upper limit of(Round to two decimal places as needed)

Answered: A clinic offers a weight loss program…

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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Identify the p-value & confidence interval
#3 Use the given confidence interval to find the margin of error and the sample proportion. (0.629,0.659) E= (Type an integer or a decimal.) p%3(Type an integer or a decimal.) Type here to search ■ 直 0 F4 F5 F7 2 4.
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…
Work Time Lost due to Accidents At a large company, the Director of Research found that the average work time lost by employees due to accidents was 92 hours per year. She used a random sample of 21 employees. The standard deviation of the sample was 5.6 hours. Estimate the population mean for the number of hours lost due to accidents for the company, using a 95% confidence interval. Assume the variable is normally distributed. Round intermediate answers to at least three decimal places. Round your final answers to the nearest whole number. 04-0 < <
n1=16 x‾1=124 s1=35 n2=12 x‾2=86 s2=31 Use this data to find the 98% confidence interval for the true difference between the population means. Assume that the population variances are not equal and that the two populations are normally distributed. Copy Data Step 2 of 3 : Find the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places. Step 3 of 3 : Please also find the high and low endpoints and round to four decimal points.
just the last part thank you!
The data below are the ages of a random sample of 8 men in a bar. It is known that the population of ages are normally distributed with a standard deviation of 7. Determine the 90% confidence interval estimate of the population mean. 49, 68, 25, 36, 56, 38, 27, 47 Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits. Confidence Interval =
Test stat: p value:
13-
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onsider the following results for two independent random samples taken from two populations. Sample 1 Sample 2 n 1 = 50 n 2 = 30 x 1 = 13.1 x 2 = 11.2 σ 1 = 2.3 σ 2 = 3 What is the point estimate of the difference between the two population means? (to 1 decimal) Provide a 90% confidence interval for the difference between the two population means (to 2 decimals). Use z-table.( , ) Provide a 95% confidence interval for the difference between the two population means (to 2 decimals). Use z-table. If your answer is negative, enter minus (-) sign.( , )
Population 2 Assume that both populations are normally distributed. (a) Test whether µ, # H, at the a = 0.05 level of significance for the given sample data. Population 1 n 16 16 (b) Construct a 95% confidence interval about u, - 42. 10.9 8.7 2.2 2.6 ...

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