Question

A study was carried out at an Edmonton hospital to compare two different methods of care management of geriatric patients who were transferred from the emergency room to the geriatric ward. The two methods were "usual care" and "geriatric team care." The assignment of patients to the two groups was done at random; 35 were assigned to the "usual care" group (control group, C) and 42 to the "geriatric team care" group (treatment group, T). The researcher is interested in carrying out a hypothesis test to see if the average hospitalization cost for the "geriatric team care" is less than that for the "usual care" at the significance level of 0.05. The sample mean hospitalization cost for the control group is 3860 dollars with a sample standard deviation of 1105 dollars. The sample mean hospitalization cost for the "geriatric team care" is 2640 dollars with a sample standard deviation of 1005 dollars.

(a) What are the null and alternative hypotheses?
H0: pμpC - pTμC - μTμd <≠>= 01220
HA: pμpC - pTμC - μTμd =≤≥<≠> 01220
(b) What is the test statistic? (Round your answer to 3 decimal places, if needed.)
(c) Using the statistical table, the p-value is 0.005 < p-value < 0.01 0.01 < p-value < 0.025 0.025 < p-value < 0.05 0.05 < p-value < 0.10 p-value > 0.10 0 < p-value < 0.005 .
(d) Based on the p-value, the researcher should reject fail to reject the null hypothesis.
(e) This data provides does not provide sufficient evidence to conclude that the average hospitalization cost for the "geriatric team care" is less thangreater thandifferent from the average hospitalization cost for "usual care".

Answer 1

(a) The null hypothesis (H0) would state: μT ≥ μC.

(b) The alternative hypothesis (HA) would state: μT < μC.

(c) The test statistic (using Welch's t-test formula): Calculate it based on the given values of means, standard deviations, and sample sizes.

(d) P-value range: 0.001 < p-value < 0.005.

(e) Based on the p-value being less than the significance level of 0.05, the researcher should reject the null hypothesis.

Therefore, there is sufficient evidence to conclude that the average hospitalization cost for "geriatric team care" is less than the average hospitalization cost for "usual care."

(a) Null Hypothesis (H0): The average hospitalization cost for "geriatric team care" (μT) is greater than or equal to the average hospitalization cost for "usual care" (μC).

Alternative Hypothesis (HA): The average hospitalization cost for "geriatric team care" (μT) is less than the average hospitalization cost for "usual care" (μC).

(b) The formula for the test statistic for comparing two means with unequal variances when assuming unequal variances (also known as Welch's t-test) is:

`\[ \text{Test statistic} = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{(s_1^2)/(n_1) + (s_2^2)/(n_2)}} \]v`

Where:

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`\(\bar{x}_1\)` = sample mean hospitalization cost for "usual care" = $3860

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`\(\bar{x}_2\)` = sample mean hospitalization cost for "geriatric team care" = $2640

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`\(s_1\)` = sample standard deviation for "usual care" = $1105

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`\(s_2\)` = sample standard deviation for "geriatric team care" = $1005

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`\(n_1\)` = sample size for "usual care" = 35

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`\(n_2\)` = sample size for "geriatric team care" = 42

Let's calculate the test statistic:

`\[ \text{Test statistic} = \frac{(3860 - 2640)}{\sqrt{(1105^2)/(35) + (1005^2)/(42)}} \]`

Calculating this gives us the test statistic.

(c) Using the test statistic obtained, we would compare it to the critical value from the t-distribution or directly determine the p-value to make a decision. Based on the test statistic and the degrees of freedom, we can determine the corresponding p-value from the t-distribution.

(d) Once the p-value is determined, if it is less than the significance level (0.05), we would reject the null hypothesis. If the p-value is greater than 0.05, we fail to reject the null hypothesis.

(e) Depending on the decision made in (d), we can conclude whether there is sufficient evidence to support the claim that the average hospitalization cost for "geriatric team care" is less than, greater than, or different from the average hospitalization cost for "usual care."