Respuesta :
Answer:
[tex]98.9-2.62\frac{0.65}{\sqrt{103}}=98.73[/tex]
[tex]98.9+2.62\frac{0.65}{\sqrt{103}}=99.07[/tex]
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".
[tex]\bar X=98.9[/tex] represent the sample mean for the sample
[tex]\mu[/tex] 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:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=103-1=102[/tex]
Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], 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 [tex]t_{\alpha/2}=2.62[/tex]
Now we have everything in order to replace into formula (1):
[tex]98.9-2.62\frac{0.65}{\sqrt{103}}=98.73[/tex]
[tex]98.9+2.62\frac{0.65}{\sqrt{103}}=99.07[/tex]
So on this case the 99% confidence interval would be given by (98.73;99.07)
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)
