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could you help me solve this exercise? It is known that the duration of treating a disease by a standard therapy has a mean of 15 days with a standard deviation of 3 days. It is claimed that a new therapy can reduce the treatment time. To test this claim, the new therapy is tried on 70 patients, and their average treatment time is calculated to be 13.6 days. Is the claim supported by this finding if the probability of Type I error is 0.01? 1. Test by using the rejection region approach. 2. Test p-value approach. 3. Is the sample in this study large enough?

User Remi Sture
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Answer:

Explanation:

To test the claim that the new therapy can reduce the treatment time, we can use a one-sample t-test.

1. Test by using the rejection region approach:

- Null hypothesis (H0): The mean treatment time using the new therapy is equal to or greater than the mean treatment time using the standard therapy (μ >= 15).

- Alternative hypothesis (Ha): The mean treatment time using the new therapy is less than the mean treatment time using the standard therapy (μ < 15).

To determine the rejection region, we need to find the critical value for a one-tailed t-test with a significance level of 0.01 and degrees of freedom n-1 (n = sample size). Since the sample size is 70, the degrees of freedom are 69.

Using a t-table or calculator, the critical value for a one-tailed t-test with 69 degrees of freedom and a significance level of 0.01 is approximately -2.649.

Next, we calculate the t-value using the formula:

t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))

t = (13.6 - 15) / (3 / sqrt(70)) = -2.711

Since the t-value (-2.711) falls in the rejection region (less than -2.649), we reject the null hypothesis. The claim that the new therapy can reduce the treatment time is supported by this finding.

2. Test p-value approach:

- Null hypothesis (H0): The mean treatment time using the new therapy is equal to or greater than the mean treatment time using the standard therapy (μ >= 15).

- Alternative hypothesis (Ha): The mean treatment time using the new therapy is less than the mean treatment time using the standard therapy (μ < 15).

To calculate the p-value, we can use the t-distribution with degrees of freedom n-1 (69 in this case) and the t-value (-2.711) obtained earlier.

Using a t-table or calculator, the p-value for a one-tailed t-test with 69 degrees of freedom and a t-value of -2.711 is approximately 0.004.

Since the p-value (0.004) is less than the significance level of 0.01, we reject the null hypothesis. The claim that the new therapy can reduce the treatment time is supported by this finding.

3. Is the sample in this study large enough?

To determine if the sample size is large enough, we can check if the sample satisfies the conditions for using the central limit theorem. The central limit theorem states that for a sample size larger than 30, the sampling distribution of the sample mean will be approximately normally distributed regardless of the shape of the population distribution.

In this case, the sample size is 70, which is larger than 30. Therefore, the sample in this study is considered large enough to apply the t-test and utilize the central limit theorem.

User HaroldSer
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