Question

**Salmon Weights:** 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 22.2 pounds with a standard deviation of
4.6 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 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

Answer:

n=26, = 22.2 , s= 4.6

c= 95%

a)

**Point estimate =
22.2**

b)

formula for confidence interval is

Where tc is the t critical value for c=95% with df=n-1 = 26-1 = 25

using t table we get critical value as

tc = 2.060

20.342 < < 24.058

round to 1 decimals

**20.3 < <
24.1**

c)

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

d)

**Because the parent
population is assumed to be normally
distributed.**

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