Question

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 19.2 pounds with a standard deviation of 4.7 pounds. You want to construct a 95% 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) Construct the 95% confidence interval for the mean weight of all spawning Chinook salmon in the Columbia River. Round your answers to 1 decimal place. < μ < (c) Are you 95% confident that the mean weight of all spawning Chinook salmon in the Columbia River is greater than 15 pounds and why? No, because 15 is above the lower limit of the confidence interval. Yes, because 15 is below the lower limit of the confidence interval. No, because 15 is below the lower limit of the confidence interval. Yes, because 15 is above the lower limit of the confidence interval. (d) Recognizing the sample size is less than 30, why could we use the above method to find the confidence interval? Because we do not know the distribution of the parent population. Because the sample size is greater than 10. Because the parent population is assumed to be normally distributed. Because the sample size is less than 100.

Answer 1

Solution:

   we are given that: The weights of spawning Chinook salmon in the Columbia river are normally distributed.

Sample Size = n = 18

Sample mean =

Samle Standard Deviation = s = 4.7

Part a) What is the point estimate for the mean weight of all spawning Chinook salmon in the Columbia River?

The point estimate for the mean weight of all spawning Chinook salmon in the Columbia River is sample mean, because sample mean is unbiased estimator of population mean and it is:

Part b) Construct the 95% confidence interval for the mean weight of all spawning Chinook salmon in the Columbia River.

Since population is normally distributed , sample size is small ( n = 18 < 30) and population standard deviation is unknown we use t distribution to find confidence interval for mean:

where

where t_c is t critical value for c = 95% confidence level

thus find two tail area = 1 - c = 1 - 0.95 = 0.05

and df = n - 1 = 18 - 1 = 17

Thus we get:

t_c = 2.110

Thus

Thus 95% confidence interval is:

Part c) Are you 95% confident that the mean weight of all spawning Chinook salmon in the Columbia River is greater than 15 pounds and why?

Yes, because 15 is below the lower limit of the confidence interval.

Part d) 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.