a.) Use a t-distribution to answer this question. Assume the samples are random samples from distributions...

a.)

Use a t-distribution to answer this question. Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used.

Find the endpoints of the t-distribution with 2.5% beyond them in each tail if the samples have sizes n1=16 and n2=29 .

Enter the exact answer for the degrees of freedom and round your answer for the endpoints to two decimal places.

b.) Use the t-distribution to find a confidence interval for a difference in means u1-u2 given the relevant sample results. Give the best estimate for u1-u2, 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 u1-u2 using the sample results x-bar1=8.5 ,s1=2.2 ,n1=50 and x-bar2=12.2 ,s2=5.4 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.

Best Estimate = Margin of error = Confidence Interval:

c.) Use a t-distribution to find a confidence interval for the difference in means ud=u1-u2 using the relevant sample results from paired data. Assume the results come from random samples from populations that are approximately normally distributed, and that differences are computed using d=x1-x2.

A 90% confidence interval for ud using the paired difference sample results x-bar of d = 564.6, sd=145.5, and nd=100

Give the best estimate for ud, the margin of error, and the confidence interval.

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.

Best Estimate = Margin of Error = Confidence Interval:

Solutions

Expert Solution

df = n1+n2 -2 = 16+29- 2 = 43

The critical value of t for 0.95 confidence interval using excel function "=TINV(0.05, 43)" is 2.017.

Answer: 2.017

orchestra answered 2 years ago

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