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Mathematics

1 answer:

Oksi-84 [34.3K]2 years ago

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Answer:

Step-by-step explanation:

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Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.â Suppose a small group of 11 Allen's humming

alexira [117]

Answer:

$3.15-1.28\frac{0.38}{\sqrt{11}}=3.003$

$3.15+ 1.28\frac{0.38}{\sqrt{11}}=3.297$

So on this case the 80% confidence interval would be given by (3.003;3.297)

And the margin of error is given by:

$ME = 1.28\frac{0.38}{\sqrt{11}}= 0.147$

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".

$\bar X=3.15$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=0.38$ represent the population standard deviation

n=11 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.80 or 80%, the value of $\alpha=0.2$ and $\alpha/2 =0.1$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NOR.INV(0.1,0,1)".And we see that $z_{\alpha/2}=1.28$