Question

Assume that the weights of spawning Chinook salmon in the Columbia river are normally distributed. You randomly catch and weigh 26 such salmon. The mean weight from your sample is 31.2 pounds with a standard deviation of 4.6 pounds. You 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) 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 < (c) Are you 90% confident that the mean weight of all spawning Chinook salmon in the Columbia River is greater than 28 pounds and why? No, because 28 is above the lower limit of the confidence interval. Yes, because 28 is below the lower limit of the confidence interval. No, because 28 is below the lower limit of the confidence interval. Yes, because 28 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 sample size is less than 100. 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.

Answer #1

Solution :

Given that,

a) Point estimate = sample mean = = 31.2

sample standard deviation = s = 4.6

sample size = n = 26

Degrees of freedom = df = n - 1 = 26 - 1 = 25

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2
= 0.05

t/2,df
= t_{0.05,25} = 1.708

Margin of error = E = t_{/2,df}
* (s /n)

= 1.708 * ( 4.6 / 26)

Margin of error = E = 1.5

The 90% confidence interval estimate of the population mean is,

- E < < + E

31.2 - 1.5 < < 31.2 + 1.5

( 29.7 < < 32.7 )

c) Yes, because 28 is below the lower limit of the confidence interval.

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

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