Question

1)

Suppose a sample of 36 patients were selected to be transferred from one hospital to a nursing home the data shown in the attached file, where the mean =17.14 and the standard deviation is calculated which equal to 0.7896. Now suppose that the health administrator wishes to construct a 95% confidence interval for the mean delay in transferring patients.

a) Find the Z-value and the Error

b) Find the confidence interval for the mean at 95%.

c) Give a small summary to the health administrator about the interval and the mean.

2) Suppose that the level of significance alpha α = 0.01 used in the sleep study. Would the evidence have been strong enough to reject the null hypothesis? (The p-value was 0.007.)

3) Suppose you want to estimate with 80% confidence interval, the population mean processing time to within ± 4 days. On the basis of the study conducted last year, you believe the standard deviation is 25 days. Determine the minimum sample size

Answer 1

A) At 95% confidence interval the critical value is z_0.025 = 1.96

Margin of error = z_0.025 * s/

                         = 1.96 * 0.7896/

                         = 0.26

b) The 95% confidence interval is

+/- ME

= 17.14 +/- 0.26

= 16.88, 17.4

c) We are 95% confident that the true population mean lies in the above confidence interval.

2) Since the P-value is less than the significance level(0.007 < 0.01), so we should reject the null hypothesis.

3) At 80% confidence interval the critical value is z_0.1 = 0.9

Margin of error = 4

or, z_0.1 * = 4

or, 1.28 * 25/ = 4

or, n = (1.28 * 25/4)^2

or, n = 64