Question

Samples from two independent, normally-distributed populations produced the following results.

Answer 1

Answer: The 95% confidence interval for the difference between population means μ1-μ2 is $(-9.36,\ 15.96)$ .

Step-by-step explanation:

Given : Samples from two independent, normally-distributed populations produced the following results.

                                     Population 1                 Population 2

Sample size                            7                                   9

Sample mean                        15.9                              12.6

Sample standard deviation 10.2                              13.4

The confidence interval for the difference between population means μ1-μ2 is given by :-

$\overline{X_1}-\overline{X_2}\pm t^*\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}$

, where $n_1$ = sample size from population 1

$n_1$  = sample size from population 2

$\overline{X_1}-\overline{X_2}$ = Difference between sample mean of two population

$s_1$= Sample standard deviation of population 1.

$s_2$= Sample standard deviation of population 2.

t* = Critical value for $df=n_1+n_2-2$ and significance $\alpha/2$.

As per given :

$n_1=7$   $n_2=9$

df = 7+9-2=14

$\overline{X_1}-\overline{X_2}=15.9-12.6=3.3$

$s_1=10.2$  $s_2=13.4$

$\alpha=1-0.95=0.05$

Critical t-value : $t_{df, \alpha/2}=t_{14, 0.025}=2.145$

So , the 95% confidence interval for the difference between population means μ1-μ2 would be

$3.3\pm (2.145)\sqrt{\dfrac{10.2^2}{7}+\dfrac{13.4^2}{9}}$

$3.3\pm (2.145)\sqrt{14.86+19.95}$

$3.3\pm (2.145)\sqrt{34.81}$

$3.3\pm (2.145)(5.9)$

$3.3\pm 12.66$

$(3.3-12.66,\ 3.3+12.66)=(-9.36,\ 15.96)$

Hence, the 95% confidence interval for the difference between population means μ1-μ2 : $(-9.36,\ 15.96)$