Question

For the following PAIRED OBSERVATIONS, calculate the 90% confidence interval for the population mean mu_d: A - 17.85, 22.56, 18.59), B-{4.95, 6.82,6.37). Yanitiniz: 0 12.25 <mu d < 14.99 O 10.72 <mu d < 16.52 10.53 <mu_d< 16.71 B.19 mu d < 19.05 o 10.47 <mu d < 16.77 13.45 <mu_d < 13.79 09.87 <mu_d< 17.37 O 10.02 <mu_d<17.22 O 13.07 <mu d < 14.17 12.27 <mu_d< 14.97 Yaniti temizle

Answer 1

The 90% confidence interval for $\mu _{d}$ is

d+tc * Sa VT

where

$\bar{d}=\frac{\sum di}{n}=40.86/3=13.62$

$s_{d}=\sqrt{\frac{\sum (di-\bar{d})^2}{(n-1)}}=\sqrt{6.9728/2}=1.8672$

degrees of freedom = n-1 = 3-1 =2

For 90% confidence interval () with df =2

a= 0.10

The two tailed critical value of t is

tc = 2.920 ( from t table)

Thus , 90% confidence interval for $\mu _{d}$ is

$13.62\pm 2.920*1.8672/\sqrt{3}$

$=13.62\pm 3.1478$

= ( 10.47 , 16.77)

Answer is

$10.47 < \mu _{d}< 16.77$

Calculation

	x1	x2	d=x1-x2	(d-dbar)^2
	17.85	4.95	12.9	0.5184
	22.56	6.82	15.74	4.4944
	18.59	6.37	12.22	1.96
sum			40.86	6.9728