In: Statistics and Probability
Salmon Weights: Assume that the weights of
spawning Chinook salmon in the Columbia river are normally
distributed. You randomly catch and weigh 25 such salmon. The mean
weight from your sample is 28.2 pounds with a standard deviation of
4.4 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 25 pounds and
why?
No, because 25 is above the lower limit of the confidence interval.Yes, because 25 is below the lower limit of the confidence interval. No, because 25 is below the lower limit of the confidence interval.Yes, because 25 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 parent population is assumed to be normally distributed.Because the sample size is greater than 10.
Solution :
Given that,
a) Point estimate = sample mean = = 28.2
sample standard deviation = s = 4.4
sample size = n = 25
Degrees of freedom = df = n - 1 = 25 - 1 =24
At 90% confidence level
= 1 - 90%
=1 - 0.90 =0.10
/2
= 0.05
t/2,df
= t0.05,24 = 1.711
Margin of error = E = t/2,df * (s /n)
= 1.711 * ( 4.4/ 25)
Margin of error = E = 1.506
The 90% confidence interval estimate of the population mean is,
- E < < + E
28.2 - 1.506 < < 28.2 + 1.506
( 26.694 < < 29.706 )
Yes, because 25 is above the lower limit of the confidence interval
d) Because the parent population is assumed to be normally distributed.
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