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 90% confidence interval for μ1-μ2 using the sample results x¯1=8.6, s1=2.7, n1=50 and x¯2=13.7, s2=5.8, n2=50 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.
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Trigonometric ratios give values of trigonometric functions. It always deals with triangles that have one angle measuring 90 degrees. These triangles are right-angled. We take the ratio of sides of these triangles.
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
A 90% confidence interval for μ1-μ2 using the sample results x¯1=8.6, s1=2.7, n1=50 and x¯2=13.7, s2=5.8, n2=50
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.
*Additional info that may be helpful: When choosing random samples of size n1 and n2 from populations with means μ1 and μ2, respectively, the distribution of the differences in the two sample means, x¯1-x¯2, has the following characteristics.
Center: The mean is equal to the difference in population means, μ1-μ2.
Spread: The standard error is estimated using SE=s12n1+s22n2.
Shape: The standardized differences in sample means follow a t-distribution with degrees of freedom approximately equal to the smaller of n1-1 and n2-1.
For small sample sizes (n1<30 or n2<30), the t-distribution is only a good approximation if the underlying population has a distribution that is approximately normal.
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