Will a random sample always accurately predict the outcome for the entire population? Why or why not? Explain your reasoning.

Mathematics

1 answer:

zvonat [6]3 years ago

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Answer:

Me lo podrías desir en español

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A data set includes 103103 body temperatures of healthy adult humans having a mean of 98.998.9degrees°f and a standard deviation

dangina [55]

Answer:

$98.9-2.62\frac{0.65}{\sqrt{103}}=98.73$

$98.9+2.62\frac{0.65}{\sqrt{103}}=99.07$

So on this case the 99% confidence interval would be given by (98.73;99.07)

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".

$\bar X=98.9$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=0.65 represent the sample standard deviation

n=103 represent the sample size

Solution to the problem

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

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

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=103-1=102$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,102)".And we see that $t_{\alpha/2}=2.62$