How Big Are Gribbles? 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 90% confidence), and a 90% 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.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X=3.1[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

s=0.72 represent the sample standard deviation

n=50 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=50-1=49[/tex]

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,49)".And we see that [tex]t_{\alpha/2}=1.677[/tex]

The margin of error is given by:

[tex] ME= 1.677 *\frac{0.72}{\sqrt{50}}= 0.2[/tex]

Now we have everything in order to replace into formula (1):

[tex]3.1-1.677\frac{0.72}{\sqrt{50}}=2.93[/tex]

[tex]3.1+1.677\frac{0.72}{\sqrt{50}}=3.27[/tex]

So on this case the 90% confidence interval would be given by (2.93;3.27)