Question

A data set includes 103103 body temperatures of healthy adult humans having a mean of 98.998.9degrees°f and a standard deviation of 0.650.65degrees°f. construct a 9999​% confidence interval estimate of the mean body temperature of all healthy humans. what does the sample suggest about the use of 98.6degrees°f as the mean body​ temperature?

Answer 1

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 $t_{\alpha/2}=2.62$

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

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

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

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

Answer 2

Answer:

99% confidence interval estimate of the mean body temperature of all healthy humans is between a lower limit of 98.73 °F and an upper limit of 99.07 °F.

The sample suggest a lower confidence interval (98.43 °F, 98.77 °F) about the use of 98.6 °F as the mean body temperature.

Step-by-step explanation:

Confidence interval = mean +/- margin of error (E)

mean = 98.9 °F

sd = 0.650 °F

n = 103

degree of freedom (df) = n - 1 = 103 - 1 = 102

confidence level (C)= 99% = 0.99

significance level = 1 - C = 1 - 0.99 = 0.01 = 1%

From the t-distribution table, t-value corresponding to 102 df and 1% significance level is 2.6251

E = t×sd/√n = 2.6251×0.65/√103 = 0.17 °F

Lower limit = mean - E = 98.9 - 0.17 = 98.73 °F

Upper limit = mean + E = 98.9 + 0.17 = 99.07 °F

99% confidence interval is (98.73 °F, 99.07 °F)

When the mean body temperature is 98.6 °F

Lower limit = mean - E = 98.6 - 0.17 = 98.43 °F

Upper limit = mean + E = 98.6 + 0.17 = 98.77 °F

99% confidence interval when the mean body temperature is 98.6 °F is (98.43 °F, 98.77 °F)