Question

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 99% confidence), and a 99% 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 margin of error = Enter your answer; margin of error The 99% confidence interval is Enter your answer; The 99%confidence interval, value 1 to Enter your answer; The 99%confidence interval, value 2

Answer 1

a) The Margin of Error = 0.26

b) The 99% Confidence Interval = ( 2.84, 3.36)

Explanation:

a) Margin of Error

The formula for Margin of Error =

z × Standard deviation/√n

From the above question

The z score for 99% confidence interval = 2.576

Standard deviation = 0.72

n = Random number of samples = 50

Margin of Error =

2.576 × 0.72/√50

= 1.85472 /√(50)

= 0.2622970178

Approximately to 2 decimal places = 0.26

b) 99% Confidence Interval

= Mean ± Margin of Error

Mean = 3.1mm

Confidence Interval = 3.1 ± 0.26

3.1 ± 0.26

= 2.84

3.1 + 0.26

= 3.36

The 99% Confidence Interval = ( 2.84, 3.36)