Question

FULL SCREEN PINTER VERSION ACKNETE Chapter 6, Section 4-CI, Exercise 191 Use the f-distribution to find a confidence interval for a difference in means , -#given the relevant sample results. Give the best estimate for Interval. Assume the results come from random samples from populations that are approximately normally distributed the margin of error, and the confidence A 95% confidence interval for Hi --using the sample results 1 - 5.7,3 = 3.1, n = 12 and >= 44,5 = 3.2.3 - 8 Enter the exact answer for the best estimate and round your answers for the margin of error and the confidence interval to two decimal places Best estimate 1.3 Margin of error the absolute tolerance is +/-0.01 Confidence interval:

Answer 1

Best estimate / point estimate for $\inline (\mu_1 - \mu_2)$ is :  $\inline (\bar{x_1}-\bar{x_2} ) = (5.7-4.4 ) = 1.3$

AT first we calculate standrd error .

$\inline SE = s_p \times \sqrt{\frac{1}{n_1}+ \frac{1}{n_2}}$

where , s_p = pooled estimate of standard deviation , which is :

$\inline \\s_p^2 = \frac{(n_1-1)s_1^2+(n_2-2)s_2^2}{n_1+n_2-2}$

$\inline \\= \frac{(11 \times 3.1^2)+(7 \times 3.2^2)}{12+8-2} \\ \\= 9.855$

So, $\inline SE= \sqrt{9.855\times(\frac{1}{12}+\frac{1}{8})} = 1.4329$

Now, Margin of error is :

$\inline ME = t_{1-\alpha/2,df} \times SE$

here alpha = 0.05

df = (12+8-2) = 18

SO, $\inline ME = t_{1-0.05/2,18} \times 1.4329 = 2.100 \times 1.4329 = 3.0091$

Confidence interval is : ( Best estimate $\inline \pm$ ME )

= (-1.7091 , 4.3091 )

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