Question

question 2
6 pts

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)).

\

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

the researchers wish to test whether there is a significant difference in the average 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.

calculate the test statistic: (round your answer to 2 decimal places)

calculate the p-value, given that the degrees of freedom are 49:

state your decision at 0.01 level of significance: (if you decide to reject the null hypothesis, enter r in the space below; otherwise, enter n.)

Explanation:

⚡ Using what you learned: Hypothesis Testing · 🆕 New: Two-Sample t-Test (Unequal Variances)
Comparing two independent group means

Step 1: Identify the given values

From the problem description and table:

• Group 1 (Albumin):
• \( n_1 = 25 \)
• \( \bar{X}_1 = 490 \)
• \( s_1 = 60 \)

• Group 2 (Polygelatin):
• \( n_2 = 14 \)
• \( \bar{X}_2 = 240 \)
• \( s_2 = 30 \)

We want to test if there is a significant difference in the average plasma volume between the two groups (a two-tailed test):

• \( H_0: \mu_1 - \mu_2 = 0 \)
• \( H_a: \mu_1 - \mu_2

eq 0 \)

Step 2: Calculate the test statistic

Since the variances are assumed to be unequal, we use the two-sample \( t \)-test with unequal variances (Welch's \( t \)-test):

\[ t = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]

Substitute the values:

\[ t = \frac{490 - 240}{\sqrt{\frac{60^2}{25} + \frac{30^2}{14}}} \]

\[ t = \frac{250}{\sqrt{\frac{3600}{25} + \frac{900}{14}}} \]

\[ t = \frac{250}{\sqrt{144 + 64.2857}} \]

\[ t = \frac{250}{\sqrt{208.2857}} \]

\[ t = \frac{250}{14.4321} \approx 17.3225 \]

Rounding to 2 decimal places:
\[ t \approx 17.32 \]

Step 3: Calculate the p-value

We are given that the degrees of freedom (\( df \)) are \( 49 \).

Since this is a two-tailed test, the \( p \)-value is:
\[ p\text{-value} = 2 \times P(T > 17.32) \]

For a \( t \)-distribution with \( df = 49 \), a \( t \)-score of \( 17.32 \) is extremely far into the tail. The probability of obtaining a value this extreme is virtually zero.

\[ p\text{-value} \approx 0.0000 \]

Rounding to 4 decimal places (or standard scientific notation):
\[ p\text{-value} < 0.0001 \text{ (or simply } 0\text{)} \]

Step 4: State the decision at the 0.01 level of significance

We compare the \( p \)-value to the significance level \( \alpha = 0.01 \):

• Since \( p\text{-value} \approx 0 < 0.01 \), we reject the null hypothesis \( H_0 \).

Therefore, the decision is to reject, which corresponds to R.

Answer:

• Calculate the test statistic: 17.32
• Calculate the p-value, given that the degrees of freedom are 49: 0.0000 (or <0.0001)
• State your decision at 0.01 level of significance: R