Question

HW 8.3 Due Thu 07/30/2020 11: Questions The body temperatures in degrees Fahrenheit of a sample of adults in one small town are: 97.8 98.1 98.3 99.8 96.5 973 97.9 96.9 99.6 Question 1 (0/1) Question 2 (0/1) Question 3 (0/1) Assume body temperatures of adults are normally distributed. Based on this data, find the 80% confidence interval of the mean body temperature of adults in the town. Enter your answer as an open-intervalle. parentheses) accurate to 3 decimal places. Assume the data is from a normally distributed population. Grade: 0/3 Print Version 80%CI= Preview TIP Enter your answer using interval notation. Example: (2,5) Use U for union to combine intervals Example: (-00,2] (4.00) Enter each value as a number (like 5, -3, 2.2172) or as a calculation (like 5/3,243, 5+4) Enter DNE for an empty set Use oo to enter Infinity Enter each value accurate to 3 decimal places

Answer 1

	Body temperature(x)	x2
	97.8	9564.84
	98.1	9623.61
	98.3	9662.89
	99.8	9960.04
	96.5	9312.25
	97.3	9467.29
	97.9	9584.41
	96.9	9389.61
	99.6	9920.16
Total	882.2	86485.1

Here

n = 9

$\overline{x}=\frac{\sum x}{n}=\frac{882.2 }{9}=98.02$

$s=\sqrt{\frac{1}{n}\left ( \sum x^2-\frac{(\sum x)^2}{n} \right )}$

(882.2) 186485.1 - 9

  

-1.11218

From standard normal table, the z-critical value for 80% confidence interval is 1.282

( or by using excel

100(1-$\alpha$ ) = 80

  $1-\alpha = 0.80$

$\alpha = 0.20$

And

$1-\frac{\alpha }{2}= 1-\frac{0.20}{2}=0.9$

Using excel formula :

=NORMSINV(0.9)

1.282

)

Now, the 80% confidence interval of population mean is,

$\overline{x}\pm Z_{\alpha /2}*\frac{s}{\sqrt{n}}$

$98.022 \pm 1.282*\frac{1.11218 }{\sqrt{9}}$

$98.022 \pm 0.475$

  

Therefore, the 80% confidence interval of the mean body temperature of adults in the town is ( 97.547, 98.497)