Mathematics, 12.12.2019 06:31 emilylizbeth12334

Adata set includes 105 body temperatures of healthy adult humans having a mean of 98.7 degrees f and a standard deviation of 0.64 degrees f.

construct a 99% confidence interval estimate of the mean body temperature of all healthy humans.

what does the sample suggest about the use of 98.6 degrees f as the mean body temperature?

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Answer from: UnusualKc5117

The 99% confidence interval would be given by (98.536;98.864)

We are 99% confident that the true mean body temperature is between (98.536;98.864)

The value 98.6 is included on the interval but the mid point for the interval is the sample mean 98.7, so for this case 98.6 would be a value to high in order to estimate the population mean, since the best estimator for the population mean is the sample mean on this case $\hat \mu =\bar X =98.7$ .

Step-by-step explanation:

1) 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".

$\bar X=98.7$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=0.64 represent the sample standard deviation

n=105 represent the sample size

Construct a 99% confidence interval estimate of the mean body temperature of all healthy humans.

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=105-1=104

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(0.005,104)".And we see that $t_{\alpha/2}=2.62$

Now we have everything in order to replace into formula (1):

$98.7-2.62\frac{0.64}{\sqrt{105}}=98.536$

$98.7+2.62\frac{0.64}{\sqrt{105}}=98.864$

So on this case the 99% confidence interval would be given by (98.536;98.864)

What does the sample suggest about the use of 98.6 degrees F as the mean body temperature?

We are 99% confident that the true mean body temperature is between (98.536;98.864)

Answer from: Quest

Answer: d. 48 step-by-step explanation: i used a calculator so i am 100% sure these are right mark as brainliest?

Answer from: Quest

answer: what are the answer choices?

step-by-step explanation:

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Adata set includes 105 body temperatures of healthy adult humans having a mean of 98.7 degrees f and...