Salmon Weights: Assume that the weights of spawning Chinook
salmon in the Columbia river are normally distributed. You randomly
catch and weigh 18 such salmon. The mean weight from your sample is
22.2 pounds with a standard deviation of 4.6 pounds. We want to
construct a 99% confidence interval for the mean weight of all
spawning Chinook salmon in the Columbia River.
(a) What is the point estimate for the mean weight of all
spawning Chinook salmon in the Columbia River?
pounds
(b) What is the critical value of t (denoted tα/2) for a 99%
confidence interval? Use the value from the table or, if using
software, round to 3 decimal places.
tα/2 =
(c) What is the margin of error (E) for a 99% confidence
interval? Round your answer to 2 decimal places.
E = pounds
(d) Construct the 99% confidence interval for the mean weight
of all spawning Chinook salmon in the Columbia River. Round your
answers to 1 decimal place.
< μ <
(e) Based on your answer to (d), are you 99% confident that
the mean weight of all spawning Chinook salmon in the Columbia
River is greater than 21 pounds and why?
No, because 21 is above the lower limit of the confidence
interval.
Yes, because 21 is below the lower limit of the confidence
interval.
No, because 21 is below the lower limit of the confidence
interval.
Yes, because 21 is above the lower limit of the confidence
interval.
(f) Recognizing the sample size is less than 30, why could we
use the above method to find the confidence interval?
Because the parent population is assumed to be normally
distributed.
Because the sample size is greater than 10.
Because the sample size is less than 100.
Because we do not know the distribution of the parent
population.