The following is the proportion of normallung use for a random sample of 17 pati

Question

The following is the proportion of normallung use for a random sample of 17 patients with acute tired lung 0.24, 0.19, 0.40, 0.23, o.30, 0.19, 0.24, o.32, 0.28, 0.24 0.18, 0.22, 0.14, 0.30, 0.07, 0.12, 0.17 Suppose that the proportion of normal lung use for acute tired lung patients has an unknown mean of and an unknown standard deviation o it is also known that tred h proportions of normal lung use are normally distributed. a) Calculate the upper 10% percentile same as 90th percentile) of the t distribution with 16 degrees of needin 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 % percentile(Same as 95th percentle) of the standard normal datribution. e) Calculate the sample standard deviation for this data? 1) 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 mwasurement Copy your R script for the above into the text box here.

Explanation / Answer

newd1 = data.frame(prp = c(0.24,0.19,0.40,0.23,0.3,0.19,0.24,0.32,0.28,0.24,0.18,0.22,0.14,0.30,0.07,0.12,0.17))

n = length(newd1$prp)

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#library(help = "stats")

## Part a - 90%ile of t-dist with 16 DoF

t90tile16 = t.test(newd1$prp, conf.level = 0.1, alternative = c("greater"))

t90tile16$conf.int[1]

## Answer: 0.251486

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## Part b - 95%ile of t-dist with 16 DoF

t95tile16 = t.test(newd1$prp, conf.level = 0.05, alternative = c("greater"))

t95tile16$conf.int[1]

## Answer: 0.2595023

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## Part c - 95%ile of t-dist with 17 DoF

### the t-statistic for DoF = 17 and confidence level = 90%, is greater than that of DoF = 16 by 0.004

### Back calculating, new bound = t95tile16$conf.int[2] + 0.004*(sd1/sqrt(n))

t95tile17 = t95tile16$conf.int[1] + 0.004*(sd1/sqrt(n))

t95tile17

## Answer: 0.2595807

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## Part d - 95%ile of normal-dist with 17 DoF

norm90ile = quantile(newd1$prp,probs = c(0.90))

norm90ile

## Answer: 0.308

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## Part e - sample mean

# var1 = var(newd1$prp)

sd1 = sqrt(var(newd1$prp))

## Answer: 0.08078657

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## Part f - sample mean

mean1 = mean(newd1$prp)

## Answer: 0.2252941

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## Part g. 90% confidence interval for population mean:-

### [ sample_mean - z-crit*sample_stdDeviation/sqrt(n), sample_mean - z-crit*sample_stdDeviation/sqrt(n) ]; Z-crit for alpha = 0.05

Zcrit = 1.64

CI_lower = mean1 - Zcrit*(sd1/sqrt(n))

CI_upper = mean1 + Zcrit*(sd1/sqrt(n))

## Answer: Confidence interval is = [0.2057005, 0.2448877]

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## Part h. 90% confidence interval for prediction of a single future weight measurement:-

### Same as the Part h)

## Answer: Confidence interval is = [0.2057005, 0.2448877]

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