Question

Salmon Weights: Assume that the weights of spawning Chinook salmon in the Columbia river are normally distributed. You randomly catch and weigh 19 such salmon. The mean weight from your sample is 19.2 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 18 pounds and why?

No, because 18 is above the lower limit of the confidence interval.

Yes, because 18 is below the lower limit of the confidence interval.

No, because 18 is below the lower limit of the confidence interval.

Yes, because 18 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 we do not know the distribution of the parent population.

Because the sample size is less than 100.

Because the sample size is greater than 10.

Because the parent population is assumed to be normally distributed.

Answer 1

a)

Point estimate for mean = sample mean, xbar = 19.2

sample standard deviation, s = 4.6
sample size, n = 19
degrees of freedom, df = n - 1 = 18

b)

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 2.878

c)

ME = tc * s/sqrt(n)
ME = 2.878 * 4.6/sqrt(19)
ME = 3.04

d)

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (19.2 - 2.878 * 4.6/sqrt(19) , 19.2 + 2.878 * 4.6/sqrt(19))
CI = (16.2 , 22.2)

e)

No, because 18 is below the lower limit of the confidence interval.

f)
Because the parent population is assumed to be normally distributed.