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Forest bathing, also called Shinrin-yoku, is the practice of taking short, leisurely walks in a forest to enhance positive health. To test the usefulness of this practice, the time forest bathing per day was recorded among eight patients with depression who subsequently showed a marked reduction in their symptoms over the last week. The data are given in the table.Time Spent in the Forest (in hours)4.34.85.13.74.35.44.04.5(a) Find the confidence limits at a 95% CI for this sample. (Round your answers to two decimal places.) (No Response) to (No Response) hours per day(b) Suppose the null hypothesis states that the average time spent forest bathing among patients is 3.5 hours per day. What would the decision be for a two-tailed hypothesis test at a 0.05 level of significance?

Question

Forest bathing, also called Shinrin-yoku, is the practice of taking short, leisurely walks in a forest to enhance positive health. To test the usefulness of this practice, the time forest bathing per day was recorded among eight patients with depression who subsequently showed a marked reduction in their symptoms over the last week. The data are given in the table.

Time Spent in the Forest (in hours)
4.3
4.8
5.1
3.7
4.3
5.4
4.0
4.5
(a) Find the confidence limits at a 95% CI for this sample. (Round your answers to two decimal places.)
(No Response) to (No Response) hours per day

(b) Suppose the null hypothesis states that the average time spent forest bathing among patients is 3.5 hours per day. What would the decision be for a two-tailed hypothesis test at a 0.05 level of significance?

check_circleAnswer
Step 1

Solution:

a. Computing 95% confidence interval for the given sample.

Since the sample size is small and the population standard deviation is unknown, the 95% confidence interval for the population mean, μ (say) is:

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Step 2

The data represents the time spent in the forest for a sample of 8 patients with depression.  Here, n=8 and α = 0.05. For the given sample, using the Excel formulas, sample mean is 4.51 and the standard deviation is 0.564.

Degrees of freedom:

Df = n – 1 = 8 – 1 = 7.

Critical value:

The critical value is obtained using the Excel formula, “=T.INV.2T(0.05, 7)”

Thus, the critical value of Student’s t is t0.025 = 2.365.

Thus, the confidence interval becomes,

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Step 3

Thus, the 95% confidence interval is (4...

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