Respuesta :
Answer:
Step-by-step explanation:
Given that a study of lobster trap placement used a 90% confidence interval to estimate the mean trap spacing (in meters) for the population of red spiny lobster fishermen fishing in the west coast.
Margin of error = 3 metres (half of width of confidence interval)
STd deviation = 11.9 metres
For 90% confidence interval we have margin of error
=[tex]1.645(\frac{11.9}{\sqrt{n} } ) = 6\\\sqrt{n} =\\3.2685[/tex]
n =10.644
=11
Using the z-distribution, it is found that 43 teams would need to be sampled.
The margin of error of a z-confidence interval is given by:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In which:
- z is the critical value.
- [tex]\sigma[/tex] is the population standard deviation.
- n is the sample size.
The first step is finding the critical value, which is z with a p-value of [tex]\frac{1 + \alpha}{2}[/tex], in which [tex]\alpha[/tex] is the confidence level.
In this problem, [tex]\alpha = 0.90[/tex], thus, z with a p-value of [tex]\frac{1 + 0.9}{2} = 0.95[/tex], which means that it is z = 1.645.
The population standard deviation is of 11.9 metres, thus [tex]\sigma = 11.9[/tex].
We want a width of 6, thus a margin of error of [tex]M = 3[/tex]. Hence, we have to solve the equation for the margin of error for n.
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]3 = 1.645\frac{11.9}{\sqrt{n}}[/tex]
[tex]3\sqrt{n} = 1.645(11.9)[/tex]
[tex]\sqrt{n} = \frac{1.645(11.9)}{3}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.645(11.9)}{3})^2[/tex]
[tex]n = 42.6[/tex]
Rounding up, 43 teams would need to be sampled.
A similar problem is given at https://brainly.com/question/25256953
