Respuesta :
Answer: (97.084, 99.116)
Step-by-step explanation:
As per given , we have
sample size :n= 106
degree of freedom: df = 105 (df=n-1)
[tex]\overline{x}=98.10\\ s=0.61[/tex]
Significance interval : [tex]\alpha: 1-0.80=0.20[/tex]
Since , population standard deviation is unknown , so we use t-test .
Using t-value table ,
[tex]t_{df,\ \alpha/2}=t_{(105,\ 0.10)}=1.290[/tex]
Now , 80% Confidence interval will be :
[tex]\overline{x}\pm t_{df,\ \alpha/2}\dfrac{s}{\sqrt{n}}[/tex]
[tex]98.10\pm (1.290)\dfrac{0.61}{\sqrt{0.6}}[/tex]
[tex]\approx98.10\pm 1.016[/tex]
[tex]=(98.10- 1.016,\ 98.10+ 1.016)=(97.084,\ 99.116)[/tex]
Thus , The 80% confidence interval estimate of the standard deviation of body temperature of all healthy humans = (97.084, 99.116).
Using the chi-square distribution, the 80% confidence interval estimate of the standard deviation of body temperature of all healthy humans is (0.5477, 0.6856).
The sample size is [tex]n = 106[/tex]
The significance level is [tex]\alpha = 1 - 0.8 = 0.2[/tex]
The estimate, which is the sample standard deviation, is of [tex]s = 0.61[/tex].
Now, we have to find the critical values for the chi-square distribution. They are:
[tex]\chi^2_{\frac{\alpha}{2},n-1} = \chi^2_{0.05,105} = 129.92[/tex]
[tex]\chi^2_{1-\frac{\alpha}{2},n-1} = \chi^2_{0.95,105} = 82.35[/tex]
The confidence interval for the population variance is:
[tex]\frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2},n-1}} < \sigma^2 < \frac{(n-1)s^2}{\chi^2_{1-\frac{\alpha}{2},n-1}}[/tex]
[tex]\frac{105(0.61)^2}{129.92} < \sigma^2 < \frac{105(0.61)^2}{82.35}[/tex]
[tex]0.3 < \sigma^2 < 0.47[/tex]
The standard deviation is the square root of the variance, thus:
[tex]\sqrt{0.3} = 0.5477, \sqrt{0.47} = 0.6856[/tex].
The 80% confidence interval estimate of the standard deviation of body temperature of all healthy humans is (0.5477, 0.6856).
A similar problem is given at https://brainly.com/question/24309021
