Question

A data 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?

Answer 1

98,536, 98.864

Explanation:

This suggest that the mean body temperature could be very possibly be 98.6 F

Answer 2

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`.

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)(s)/(√(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(0.64)/(√(105))=98.536`

`98.7+2.62(0.64)/(√(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)

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`.