Question

Consider the folloing data for two independent random samples taken from two normal populations with equal variances. find the 80% confidence interval for µ1 - µ2. sample 1: 12,8,11,6,13,7 sample 2: 13,16,10,9,13,14 what is the left endpoint and right endpoint?

Answer 1

Confidence Interval for Difference in two Population means $\small \mu _{1}-\mu _{2}$

Formula for Confidence Interval for Difference in two Population means when population Standard deviation are equal and not known

$\small (\overline{x_{1}}-\overline{x_{2}})\pm t_{\alpha /2,n-1}\times s_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}$

$\small Pooled\; \; Sample\; \; Standard\; deviation\; \; :s_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}}$

$\small n_{1} :$ Sample Size of Sample 1

$\small n_{2} :$ Sample Size of Sample 2

$\small \overline{x_{1}}$ : Sample Mean of Sample 1

$\small \overline{x_{2}}$ Sample Mean of Sample 2

s1 : Sample Standard Deviation of Sample 1

s2 : Sample Standard Deviation of Sample 2

For Sample 1

Sample 1	x1	x1- $\small \overline{x_{1}}$	(x1- $\small \overline{x_{1}}$ )2
	12	2.5	6.25
	8	-1.5	2.25
	11	1.5	2.25
	6	-3.5	12.25
	13	3.5	12.25
	7	-2.5	6.25
Total	57		41.5
Mean: $\small \overline{x_{1}}$ :	57/6=9.5

$\small s_{1}=\sqrt{\sum \frac{(x_{1}-\overline{x}_{1})^{2}}{n_{1}-1}}=\sqrt{\frac{41.5}{6-1}}=2.881$

For Sample 2

Sample 2	x2	x2- $\small \overline{x}_{2}$	(x2- $\small \overline{x}_{2}$ )
	13	0.5	0.25
	16	3.5	12.25
	10	-2.5	6.25
	9	-3.5	12.25
	13	0.5	0.25
	14	1.5	2.25
Total	75		33.5
Mean	75/6=12.5

$\small s_{2}=\sqrt{\sum \frac{(x_{1}-\overline{x}_{1})^{2}}{n_{1}-1}}=\sqrt{\frac{33.5}{6-1}}=2.588$

n1 : Sample Size of Sample 1	6
n2 : Sample Size of Sample 2	6
$\small \overline{x_{1}}$ : Sample Mean of Sample 1	9.5
$\small \overline{x_{2}}$ : Sample Mean of Sample 2	12.5
s1 : Sample Standard Deviation of Sample 1	2.881
s2 : Sample Standard Deviation of Sample 2	2.588
Confidence Level	80%
$\small \alpha$ (= 100-80/100=20/100 ) = 0.2	0.2
$\small \alpha$ /2 (=0.2/2=0.1)	0.1
Degrees of freedom : ( n1+n2-2=6+6-2=10)	10
$\small t_{\alpha /2,df}=t_{0.1,10}$	1.3722

$\small Pooled\; \; Sample\; \; Standard\; deviation\; \; :s_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}}$

$\small s_{p}= \sqrt{\frac{(6-1)2.881^{2}+(6-1)2.588^{2}}{6+6-2}}=\sqrt{\frac{74.9895}{10}}=\sqrt{7.499}=2.7384$

Confidence Interval for Difference in two Population means

$\small (\overline{x_{1}}-\overline{x_{2}})\pm t_{\alpha /2,n-1}\times s_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}$

$\small =(9.5-12.5)\pm1.3722\times2.7384 \sqrt{\frac{1}{6}+\frac{1}{6}}$

$\small =-3\pm1.3722\times2.7384\times0.5774$

$\small =-3\pm2.1695=(-5.1695,-0.8305)$

80% confidence interval for µ1 - µ2 = (-5.1695,-0.8305)

left endpoint : -5.1695

right endpoint : -0.8305