The following is the proportion of normal lung use for a random sample of 17 pat

Question

The following is the proportion of normal lung use for a random sample of 17 patients with acute tired lung 0.24, 0.19, 0.40, 0.23, 0.30, 0.19, 0.24, 0.32, 0.28, 0.24 0.18, 0.22, 0.14, 0.30, 0.07, 0.12, 0.17 Suppose that the proportion o normal ung use or acute tired ung patients has an unknown mean o and an unknown standard dev a ono. t sa o now tha tire un proportions or norma un use are normally distributed. a) Calculate the upper 10% percentile(Same as 90th percentile) of the t distribution with 16 degrees of freedom. b) Calculate the upper 5% percentile(Same as 95th percentile) of the t distribution with 16 degrees of freedom c) Calculate the upper 5% percentile(Same as 95th percentile) of the t distribution with 17 degrees of freedom. d) Calculate the upper 5% percentile(Same as 95th percentile) of the standard normal distribution. e) Calculate the sample standard deviation for this data? f) Calculate the sample mean for this data g) Compute a 90% Confidence Interval for . h)Compute a 90% Prediction Interval for a single future weight measurement.

Explanation / Answer

xbar 0.2253 std. dev. 0.0808 n 17 SE = s/sqrt(n) 0.0196 A. t-value for upper 90% 1.3368 x = xbar + t*SE 0.2515 B. t-value for upper 95% 1.7459 x = xbar + t*SE 0.2595 C. t-value for upper 95%, df = 17 1.7396 x = xbar + t*SE 0.2594 D. z-value for upper 95% 1.6449 x = xbar + z*SE 0.2575 E. s 0.0808 F. mean 0.2253 G. z-value for 90%CI 1.6449 ME = z*SE 0.0322 Lower Limit = mean - ME 0.1931 Upper Limit = mean + ME 0.2575

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