Question

after surgery a patients blood volume is often depleted. in one study, the total circulating volume of blood plasma was measured for each patient immediately after surgery. after infusion of a \plasma expander\ into the bloodstream, the plasma volume was measured again and the increase in plasma volume (ml) was calculated. two of the plasma expanders used were albumin (25 patients) and polygelatin (14 patients). the following table reports the increase in plasma volume (samuels, witmer, and schaffner (1989)).

\

$$\begin{tabular}{|c|c|c|} \\hline & albumin & polygelatin \\\\ \\hline n & 25 & 14 \\\\ \\hline \\bar{x} & 490 & 240 \\\\ \\hline s & 60 & 30 \\\\ \\hline \\end{tabular}$$

research wish to construct a 99\\% confidence interval (ci) to estimate the difference in the mean plasma volume between the two types of plasma expanders. assume that the data from both groups are approximately normally distributed and that the variances are different across the two groups. compute the following.

compute the point estimate:

compute the critical value (df=49): (round your answer to 2 decimal places)

compute the standard error: (round your answer to 2 decimal places)

lower limit of the ci: (round your answer to 4 decimal places)

upper limit of the ci: (round your answer to 4 decimal places)

Explanation:

⚡ Using what you learned: Point Estimation and Interval Estimation

Step 1: Compute the point estimate

The point estimate for the difference between the two population means, \( \mu_1 - \mu_2 \), is the difference between the sample means:

$$ \text{Point Estimate} = \bar{X}_1 - \bar{X}_2 $$

$$ \text{Point Estimate} = 490 - 240 = 250 $$

Step 2: Compute the critical value

For a \( 99\% \) confidence interval with degrees of freedom \( \text{df} = 49 \), we find the critical value \( t^* \) (or \( t_{\alpha/2} \)) using a \( t \)-distribution table or calculator:

$$ \alpha = 1 - 0.99 = 0.01 \implies \frac{\alpha}{2} = 0.005 $$

$$ t_{0.005, \, 49} \approx 2.68 $$

Step 3: Compute the standard error

Since the variances are assumed to be different, we use the unpooled standard error formula for two independent samples:

$$ \text{SE} = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} $$

$$ \text{SE} = \sqrt{\frac{60^2}{25} + \frac{30^2}{14}} = \sqrt{\frac{3600}{25} + \frac{900}{14}} $$

$$ \text{SE} = \sqrt{144 + 64.285714} = \sqrt{208.285714} \approx 14.4321 $$

Rounding to 2 decimal places:

$$ \text{SE} \approx 14.43 $$

Step 4: Compute the margin of error

Using the unrounded values for precision:

$$ \text{ME} = t^* \times \text{SE} = 2.67995 \times 14.43211 \approx 38.6773 $$

Step 5: Compute the confidence interval limits

$$ \text{Lower Limit} = \text{Point Estimate} - \text{ME} = 250 - 38.6773 = 211.3227 $$

$$ \text{Upper Limit} = \text{Point Estimate} + \text{ME} = 250 + 38.6773 = 288.6773 $$

Answer:

• Compute the point estimate: 250
• Compute the critical value (df=49): 2.68
• Compute the standard error: 14.43
• Lower Limit of the CI: 211.3227
• Upper Limit of the CI: 288.6773