The ratio of catfish to trout in the local pond is 2 to 5. If there are 300 catfish in the pond, how many trout are there?
1 answer:
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The approximate volume of the model will be
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Answer:
a) The 99% confidence interval would be given by (589.588;731.038)
b) If we see the confidence interval the 620 is included on the interval we don't have enough evidence to reject the rating is 620. But since we need a score that at least 620 and our lower limit is 589.588 we cant conclude that all the clients would have a score that at least of 620.
Step-by-step explanation:
Previous concepts
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".
The data is:
661 595 548 730 791 678 672 491 492 583 762 624 769 729 734 706
Part a
Compute the sample mean and sample standard deviation.
In order to calculate the mean and the sample deviation we need to have on mind the following formulas:
=AVERAGE(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)
On this case the average is
=STDEV.S(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)
The sample standard deviation obtained was s=95.898
Find the critical value t* Use the formula for a CI to find upper and lower endpoints
In order to find the critical value we need to take in count that our sample size n =16<30 and on this case 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 99% of confidence, our significance level would be given by and
. The degrees of freedom are given by:
We can find the critical values in excel using the following formulas:
"=T.INV(0.005,15)" for
"=T.INV(1-0.005,15)" for
The confidence interval for the mean is given by the following formula:
And if we find the limits we got:
[tex]660.313+ 2.95\frac{95.898}{\sqrt{16}}=731.038/tex]
So the 99% confidence interval would be given by (589.588;731.038)
Part b
If we see the confidence interval the 620 is included on the interval we don't have enough evidence to reject the rating is 620. But since we need a score that at least 620 and our lower limit is 589.588 we cant conclude that all the clients would have a score that at least of 620.