Mathematics, 18.12.2019 03:31, JustTraaash

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 19.2pounds 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.
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(c) are you 90% 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 sample size is greater than 10.

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 less than 100.

Answers: 1

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Answers

a) $\bar X= 19.2$ represent the sample mean. And that represent the best estimator for the population mean since $\hat \mu =\bar X=19.2$ b) The 90% confidence interval is given by (17.3;21.1)

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

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

The reason of this is because the t distribution is an special case of the normal distribution when the degrees of freedom increase.

Step-by-step explanation:

1) Notation and definitions

n=17 represent the sample size

Part a

$\bar X= 19.2$ represent the sample mean. And that represent the best estimator for the population mean since $\hat \mu =\bar X=19.2$ Part b

s=4.4 represent the sample standard deviation

m represent the margin of error

Confidence =90% or 0.90

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

2) Calculate the critical value tc

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$ . The degrees of freedom are given by:

df=n-1=17-1=16

We can find the critical values in excel using the following formulas:

"=T.INV(0.05,16)" for $t_{\alpha/2}=-1.75$

"=T.INV(1-0.05,16)" for $t_{1-\alpha/2}=1.75$

The critical value $tc=\pm 1.75$

3) Calculate the margin of error (m)

The margin of error for the sample mean is given by this formula:

$m=t_c \frac{s}{\sqrt{n}}$

$m=1.75 \frac{4.4}{\sqrt{17}}=1.868$

4) Calculate the confidence interval

The interval for the mean is given by this formula:

$\bar X \pm t_{c} \frac{s}{\sqrt{n}}$

And calculating the limits we got:

$19.2 - 1.75 \frac{4.4}{\sqrt{17}}=17.332$

$19.2 + 1.75 \frac{4.4}{\sqrt{17}}=21.068$

The 90% confidence interval is given by (17.332;21.068) and rounded would be: (17.3;21.1)

Part c

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

Part d

Because the parent population is assumed to be normally distributed.

The reason of this is because the t distribution is an special case of the normal distribution when the degrees of freedom increase.

answered: Guest

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