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 99% confidence interval for μ1-μ2 using the sample results x¯1=507, s1=112, n1=360 and x¯2=445, s2=96, n2=200 Enter the exact answer for the best estimate and round your answers for the margin of error and the confidence interval to two decimal places.

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=

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

You are going to create a 99% confidence interval for the mean amount ofsugar in coffee shop specialty drinks. You know the population standarddeviation is σ = 5.9 grams. You want your confidence interval to have amargin of error of 0.25 grams. How big should your sample size be to obtainthis margin of error?

A researcher collected sample data for 10 women ages 18 to 24. The sample had a mean serum cholesterol level (measured in mg/100 mL) of 188.4, with a standard deviation of 5.7. Assuming that serum cholesterol levels for women ages 18 to 24 are normally distributed, find a 99% confidence interval for the mean serum cholesterol level of all women in this age group. Then find the lower limit and upper limit of the  99%  confidence interval.

Assume that a sample is used to estimate a population mean μ . Find the 95% confidence interval for a sample of size 582 with a mean of 34.8 and a standard deviation of 8.1. Enter your answers accurate to three decimal places. Confidence Interval = (        ,       )