Question

Colors Fonts Set as Default Document Formatting Answer the following questions: Question 1. [3+5=8 Marks] Random samples have been selected from two normally distributed populations with equal variances. The following sample data were observed, Samples from population 1 Samples from population 2 a) Find a 90% confidence interval for the difference between two population means. b) Use a significance level of 0.10 for testing the hypothesis 11 HH2 10. (United States)

Answer 1

Q1.

From data we get : n1 = 7, n2 = 7, $\overline{x}_{1}$ = 14.14, $\overline{x}_{2}$ = 11.57, s₁ = 1.345, s₂ = 1.618

Here population variances are equal .

Hence 90% confidence interval is given by,

{ ( $\overline{x}_{1}$ - $\overline{x}_{2}$ ) - E, ( $\overline{x}_{1}$ - $\overline{x}_{2}$ ) + E }

Where,

$E=tc*s*\sqrt{\frac{1}{n1}+\frac{1}{n2}}$

Where,

$s=\sqrt{\frac{(n1-1)*s_{1}^{2}+(n2-1)*s_{2}^{2}}{n1+n2-2}}$

$=\sqrt{\frac{(7-1)*1.345^{2}+(7-1)*1.618^{2}}{7+7-2}}$

= 1.488

c = 0.90, df = n1+n2-2 = 7+7 -2 = 12 , $\alpha =1-c=1-0.90=0.10$

Hence t_c = $t_{\alpha /2,df}=t_{0.10/2,12}=1.782$ ------using Excel formula "=t.inv.2t(0.10,12)"  

Hence the margin of error is,

$E=1.782*1.488*\sqrt{\frac{1}{7}+\frac{1}{7}}$

= 1.42

Hence the 90% confidence interval is,

{ ( 14.14 -11.57) -1.42 , ( 14.14 -11.57 ) + 1.42 }

( 1.15, 3.99 )

b) The hypothesis are

H0: $\mu _{1}=\mu _{2}$ v/s H1: $\mu _{1}\neq \mu _{2}$

The test statistic is,

$t=\frac{\overline{x}_{1}-\overline{x}_{2}}{s*\sqrt{\frac{1}{n1}+\frac{1}{n2}}}$

$=\frac{14.14-11.57}{1.488*\sqrt{\frac{1}{7}+\frac{1}{7}}}$

= 1.814

The critical value is given by,

df = n1+n2 -2 = 7+7 -2 = 12 , $\alpha$ = 0.10

Hence the critical value = $t_{\alpha /2,df}=t_{0.10/2,12}=\pm 1.782$ ------using Excel formula "=t.inv.2t(0.10,12)"  

So the calculated value of t > critical value of t .

Hence we reject the null hypothesis.

Conclusion :

There is difference between two population means.