Question

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 4.8 pounds. We want to construct a 90% 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 t_α/2) for a 90% confidence interval? Use the value from the table or, if using software, round to 3 decimal places.
t_α/2 =

(c) What is the margin of error (E) for a 90% confidence interval? Round your answer to 2 decimal places.
E =  pounds

(d) Construct the 90% 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 90% confident that the mean weight of all spawning Chinook salmon in the Columbia River is greater than 19 pounds and why?

No, because 19 is above the lower limit of the confidence interval.

Yes, because 19 is below the lower limit of the confidence interval.    

No, because 19 is below the lower limit of the confidence interval.

Yes, because 19 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.

Answer 1

a) The point estimate for the mean weight of all spawning Chinook salmon in the Columbia River
22.2 pounds.

(b) The critical value of t (denoted t_α/2) for a 90% confidence interval
t_α/2 =1.740

(c) The margin of error (E) for a 90% confidence interval?
E =1.97 pounds

(d) Construct the 90% confidence interval for the mean weight of all spawning Chinook salmon in the Columbia River.

μ = M ± t(s_M)
μ = 22.2 ± 1.74*1.13
μ = 22.2 ± 1.97
20.23 < μ <24.17

(e) Based on your answer to (d), are you 90% confident that the mean weight of all spawning Chinook salmon in the Columbia River is greater than 19 pounds and why?

Yes, because 19 is below 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.