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?

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".

represent the sample mean for the sample  

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:

  (1)

In order to calculate the critical value we need to find first the degrees of freedom, given by:

Since the Confidence is 0.99 or 99%, the value of and , 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

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

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

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)

took the test it's 94

its a trick question

step-by-step explanation:

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