Question

Salmon Weights: Assume that the weights of spawning Chinook salmon in the Columbia river are normally distributed. You randomly catch and weigh 17 such salmon. The mean weight from your sample is 22.2 pounds with a standard deviation of 4.4 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 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.

(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.Because the sample size is greater than 10.     Because we do not know the distribution of the parent population.Because the sample size is less than 100.

Answer 1

(a) point estimate for the mean weight of all spawning Chinook salmon in the Columbia River=22.2

sample mean is the unbiased estimated of the population mean, so sample mean would be the point estimate of the population mean

(b)95% confidence interval for population mean=(19.938,24.462)

since sample standard deviation is given/available so we use t-value to find the confidence interval

(1- α)*100% confidence interval for population mean=sample mean±t(α/2,n-1)*s/sqrt(n)

95% confidence interval =22.2±t(0.05/2, n-1)*4.4/sqrt(17)=22.2±2.120*4.4/sqrt(17)=22.2±2.262=(19.938,24.462)

	t-value	margin of error	lower limit	upper limit
95% confidence interval	2.120	2.262	19.938	24.462

n=	17
sample mean=	22.200
s=	4.400

(c) Yes, because 18 is above the lower limit of the confidence interval.

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

we use t-value assuming the parent population is normally distributed