Question

none of my answers are working, please help.

Chapter 6, Section 2-CI, Exercise 082 Your answer is partially correct. Try again. Use the t-distribution to find a confidence interval for a mean given the relevant sample results. Give the best point estimate for y, the margin of error, and the confidence interval. Assume the results come from a random sample from a population that is approximately normally distributed. A 95% confidence interval for u using the sample results ī= 75.3, s = 8.0, and n= 42 Round your answer for the point estimate to one decimal place, and your answers for the margin of error and the confidence interval to two decimal places. point estimate = 175.3 margin of error = ... 177.719 The 95% confidence interval is 72.881 to Click if you would like to Show Work for this question: Open Show Work

Answer 1

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Here, given that, X= 75.3,5= 8.0, n=42 point estimate = 175.3 margin of error tn-1;4/2 2 1 where, tnijelz= tez-1; 4/2= taljoos- : 2.02 X 8.0 542 - 2.49 :. Margin of error = 12.49 gSt. (.L. for et :- (x - margin of (75.3-2.49 error X + margin of error) 9 75.3 + 2.69 ) 9 172.81 77.79 ) The gst. confidence interval is 72.81 to 127.79