Use the t-distribution to find a confidence interval for a difference in means μ₁ −µ₂ given the relevant sample results. Give the bestestimate for μ₁ −μ₂, the margin of error, and the confidence interval. Assume the results come from random samples frompopulations that are approximately normally distributed.A 95% confidence interval for μ₁ − µ₂ using the sample results +₁ =Best estimate = iMargin of error =Confidence interval: i511, $1Enter the exact answer for the best estimate and round your answers for the margin of error and the confidence interval to twodecimal places.to i=122, n₁ = 320 and +₂ = 448, s₂= 85, n₂ = 200

Given sample 1: n₁ = 320 (sample size) x̄₁ = 511 (sample mean) s₁ = 122 (sample standard…

Answered: Use the t-distribution to find a…

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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Given two independent random samples with the following results: n1=7x‾1=179s1=22n1=7x‾1=179s1=22 n2=14x‾2=145s2=20n2=14x‾2=145s2=20 Use this data to find the 80% 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 1 of 3 : Find the point estimate that should be used in constructing the confidence interval. 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 of3: construct the 80% confidence interval.Round your answers to nearest whole number.
Use the t-distribution to find a confidence interval for a difference in means μ1-μ2 given the relevant sample results. Give the best estimate for μ1-μ2, the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed.A 95% confidence interval for μ1-μ2 using the sample results x̅1 = 75.5, s1=11.5, n1=35 and x̅2=69.3, s2=7.0, n2=20Enter the exact answer for the best estimate and round your answers for the margin of error and the confidence interval to two decimal places Best estimate =___________ Margin of error =______________ Confidence interval :_____________ to _______________
A simple random sample of 10 subjects finds that the sample mean is 56 and the standard deviation is 4. Find the 98% confidence interval for u. Display the result in interval notation. You must round your result to decimal places. Remember that the information you have is from a small sample, you must use the t distribution. Remember that critical values can be calculated in EXCEL using the degrees of freedom (n-1) and the level of significance (I - C) as a function of T.INV 2T (Level of Significance, Degrees of Freedom) Select one: a. The interval is: (53,057, 58,943) b. The range is: (53,521, 58,479). c. The interval is: (51,080, 60,920) d. The interval is: (51.991, 60.009) e. The interval is: (52,431, 59,569)
Find the critical value tcfor the confidence level c=0.80 and sample size n=28.
Age Differences In a large hospital, a nursing director selected a random sample of 43 registered nurses and found that the mean of their ages was 31.5. The population standard deviation for the ages is 4.5. She selected a random sample of 32 nursing assistants and found the mean of their ages was 29.7. The population standard deviation of the ages for the assistants is 5.3. Find the 95% confidence interval of the differences in the ages. Use μ1 for the mean age of the nursing assistants. Round the answers to at least one decimal place. Use a TI-83 Plus/TI-84 Plus calculator. ? < μ1- μ2 < ?
A simple random sample of 60 items resulted in a sample mean of 71. The population standard deviation is 13. a. Compute the 95% confidence interval for the population mean (to 1 decimal). Assume that the same sample mean was obtained from a sample of 120 items. Provide a 95% confidence interval for the population mean (to 2 decimals).
A random sample of size 40 have been selected from a normally distributed population. If the sample mean equal to 40 and the population standard deviation equal to 7. Find 95 % Confidence Interval for the Population Mean.
The 82.9% confidence interval estimate for μμ was (25.81, 31.35). Find the point estimate and the margin of error (ME) from this interval estimate. Point estimate = ME=
COMPUTE FOR THE MARGIN OF ERROR OF THE FOLLOWING AND SHOW THE STEP-BY-STEP SOLUTION. 1. Confidence level = 99%, standard deviation (s) = 25, sample size (n) = 300 FILL IN THE TABLE WITH SOLUTIONS. STEPS SOLUTION STEP 1: STEP 2: STEP 3: STEP 4: STEP 5: Margin of Error: Conclusion:
Use the t-distribution to find a confidence interval for a difference in means μ1−μ2μ1-μ2 given the relevant sample results. Give the best estimate for μ1−μ2μ1-μ2, the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed.A 95% confidence interval for μ1−μ2μ1-μ2 using the sample results x⎯⎯1=515x¯1=515, s1=136s1=136, n1=300n1=300 and x⎯⎯2=462x¯2=462, s2=94s2=94, n2=200n2=200Enter the exact answer for the best estimate and round your answers for the margin of error and the confidence interval to two decimal places.Best estimate = Enter your answer; Best estimateMargin of error = Enter your answer; Margin of errorConfidence interval : Enter your answer; Confidence interval, value 1 to Enter your answer; Confidence interval, value 2
Find the critical value x²R corresponding to a sample size of 19 and a confidence level of 99 percent. X²R = (Round to 3 decimals)
find out the confidence interval for population proportion for given data [ 4 decimal places ] population standard deviation =4 xbar = 123.5 sample size= 100 c= 0.95

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