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. 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. Interpret the confidence interval in context. What do we have to assume about the sample in order to have confidence in our estimate?

Answer 1

Point estimate = Mean = 3.1

Sample size (n) = 50

Standard deviation (s) = 0.72

Confidence interval(in %) = 95

z @ 95% = 1.96

Since we know that

Required confidence interval = (3.1-0.1996, 3.1+0.1996)

Required confidence interval = (2.9004, 3.2996)

With 95% confidence, we can say that the true mean of length of gribbles is in between 2.9004 and 3.2996

We have to assume the z distribution in order to calculate the confidence interval since sample size is greater than 30

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