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patients recovering from an appendix operation normally spend an average of 6.3 days in the hospital. the distribution of recovery times is normal with a mean equal to 1.2 days. The hospital is trying a new recovery program designed to lessen the time patients spend in the hospital. The first 10 appendix patients in this new program were released from the hospital in an average of 5.5 days. On the basis of these data, can the hospital conclude that the new program has a significant reduction recovery time. Test at the .05 level of significance.

User Abuarif
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On the basis of these data and since our calculated P-value is less than 0.05, there is sufficient evidence to support the claim that the new program significantly reduces the recovery time.

Average time spent by patients in the hospital recovering fromm an appendix operation = 6.3 days

The sample mean = 1.2 days

The average time spent by 10 patients using the new recovery program = 5.5. days

Level of significance = 0.05

This is a case of a one-sample t-test. We compare the sample mean (5.5 days) to the population mean (6.3 days).

The null hypothesis (H₀) and alternative hypothesis (Hₐ) are as follows:

Null Hypothesis (H₀): The mean recovery time under the new program is equal to 6.3 days.

Alternative Hypothesis (Hₐ): The mean recovery time under the new program is less than 6.3 days.

The test statistic for a one-sample t-test is calculated as follows:


t = (x - u)/((s)/( √(n) ))

where:


\bar x is the sample mean,

μ is the population mean,

s is the standard deviation of the sample, and

n is the sample size.

Substituting the values into the formula:


t = (5.5 - 6.3)/((2)/( √(10) )) = 2.12

Thus, the test statistic is 2.12.

The degrees of freedom in this case is 10 - 1 = 9.

The P-value associated with this t-statistic can be found using a t-distribution table or a statistical software and given a t-statistic of -2.12 with 9 degrees of freedom and a one-tailed test, the P-value would be less than 0.05.

If the P-value is less than the significance level (0.05), we reject the null hypothesis.

Thus, since our calculated P-value is less than 0.05, there is sufficient evidence to support the claim that the new program significantly reduces the recovery time.

User Crouching Kitten
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