Q:

Gribbles are small, pale white, marine worms that bore through wood. While sometimes considered a pest since they can wreck wooden docks and piers, they are now being studied to determine whether the enzyme they secrete will allow us to turn inedible wood and plant waste into biofuel.1 A sample of 50 gribbles finds an average length of 3.1 mm with a standard deviation of 0.72. Give a best estimate for the length of gribbles, a margin of error for this estimate (with 95% confidence), and a 95% confidence interval. 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 = Enter your answer; point estimate

Accepted Solution

A:
Answer:a) The Margin of Error = 0.26b) The 99% Confidence Interval = ( 2.84, 3.36)Step-by-step explanation:a) Margin of ErrorThe formula for Margin of Error =z × Standard deviation/√nFrom the above questionThe z score for 99% confidence interval = 2.576Standard deviation = 0.72n = Random number of samples = 50Margin of Error =2.576 × 0.72/√50= 1.85472 /√(50)= 0.2622970178Approximately to 2 decimal places = 0.26b) 99% Confidence Interval= Mean ± Margin of ErrorMean = 3.1mmConfidence Interval = 3.1 ± 0.263.1 ± 0.26= 2.843.1 + 0.26= 3.36The 99% Confidence Interval = ( 2.84, 3.36)