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 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

Answer 1

Answer with explanation:

As per given , we have

A sample of 50 gribbles finds an average length of 3.1 mm with a standard deviation of 0.72.

i.e. n= 50

Degree of freedom = n-1=49

$\overline{x}=3.1\ mm$

$s=0.72\ mm$

The point estimate of the population mean is equals to the sample mean.

i.e. The best estimate for the length of gribbles= 3.1 mm

Also, population standard deviation is unknown , then the confidence interval would be :

$E=t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}$

For df= 49 and significance level of 0.05 , the t- value (using t-distribution table) would be :

$t_{0.05/2, 49}=t_{0.025, 49}= 2.010$

Now, Margin of error : $E=(2.010)\dfrac{0.72}{\sqrt{50}}$

$=0.204664986747\approx0.20$

95% confidence interval : $(\overline{x}-E,\ \overline{x}+E)$

$i.e.\ (3.1-0.20, 3.1+0.20)=(2.90,\ 3.30)$