Respuesta :
Answer:
One can be 99% confident the true mean shell length lies within the above interval.
The population has a relative frequency distribution that is approximately normal.
Step-by-step explanation:
We are given that Time-depth recorders were deployed on 6 of the 76 captured turtles. These 6 turtles had a mean shell length of 51.3 cm and a standard deviation of 6.6 cm.
The pivotal quantity for a 99% confidence interval for the true mean shell length is given by;
P.Q. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean shell length = 51.3 cm
s = sample standard deviation = 6.6 cm
n = sample of turtles = 6
[tex]\mu[/tex] = true mean shell length
Now, the 99% confidence interval for [tex]\mu[/tex] = [tex]\bar X \pm t_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }[/tex]
Here, [tex]\alpha[/tex] = 1% so [tex](\frac{\alpha}{2})[/tex] = 0.5%. So, the critical value of t at 0.5% significance level and 5 (6-1) degree of freedom is 4.032.
So, 99% confidence interval for [tex]\mu[/tex] = [tex]51.3 \pm 4.032 \times \frac{6.6}{\sqrt{6} }[/tex]
= [51.3 - 10.864 , 51.3 + 10.864]
= [40.44 cm, 62.16 cm]
The interpretation of the above result is that we are 99% confident that the true mean shell length lie within the above interval of [40.44 cm, 62.16 cm].
The assumption about the distribution of shell lengths must be true in order for the confidence interval, part a, to be valid is that;
C. The population has a relative frequency distribution that is approximately normal.
This assumption is reasonably satisfied as the data comes from the whole 76 turtles and also we don't know about population standard deviation.
