Question

Please focus on f)

4. A commonly used drug produces increases in blood pressure that are normally distributed with mean μ-6 and standard deviation σ-2.5. A new drug is expected to produce normally distributed changes but with possibly different values of μ and σ. The new drug was given to 10 individuals with the following results: 8.05, 12.16, 11.16, 11.28, 11.11, 9.66, 8.40, 13.30, 5.32, 8.87 Note that9.931, 2.342 (a) Use the t-test to determine if there is evidence that μ yi 6 at level α = 0.05? Clearly state your hypotheses. (b) Give a 90% two-sided confidence interval for μ. (c) At the 5% level is there evidence that σ 2.5? (d) Use the sign test to test the hypothesis that the new drug has a higher median increase in blood pressure. Use α-0.05. (e) Use the Wilcoxon test to test the hypothesis that the new drug has a higher median increase in blood pressure. Use a 0.05 and the normal approximation to the distribution of the Wilcoxon statistic (f) Give an approximate 90% distribution-free confidence interval for the median C4] change in blood pressure. Some R output that may be useful: pec(0.1,0.05,0.025,0.01) qt(1-p,8) [1] 1.396815 1.859548 2.306004 2.896459 > qt (1-p,9) [1] 1.383029 1.833113 2.262157 2.821438 qt(1-p,10) 1] 1.372184 1.812461 2.228139 2.763769 [1] 3.489539 2.732637 2.179731 1.646497 [1) 4.168159 3.325113 2.700389 2.087901 1] 4.865182 3.940299 3.246973 2.558212 qchisq(p,8) qchisq(p,9) qchisq(P,10) qchisq(1-p,9) [1) 14.68366 16.91898 19.02277 21.66599 > qchisg(1-р, 10) [1) 15.98718 18.30704 20.48318 23.20925 round (dbinom(O:10,10,1/2),digits-2) [1] 0.00 0.01 0.04 0.12 0.21 0.25 0.21 0.12 0.04 0.01 0.00 norm(1-p [1] 1.281552 1.644854 1.959964 2.326348

Answer 1

(f)

We will calculate the confidence interval of median by bootstrap samples. Let M = 1000 be the number of bootstrapped samples.

Below is the R code:

#vector
x <- c(8.05, 12.16, 11.16, 11.28, 11.11, 9.66, 8.4, 13.3, 5.32, 8.87)
alpha <- 0.1
df = length(x) - 1

#1000 bootstrapped samples
median_vec <- c()
for (i in 1:1000) {
boot_x <- sample(x, replace = TRUE)
median_vec <- c(median_vec, median(boot_x))
  
}

#standard error of median calculation
sd_median <- sd(median_vec)

#90% confidence interval calculation
margin_of_err <- qt(1 - alpha/2, df) * sd_median

left <- median(x) - margin_of_err
right <- median(x) + margin_of_err

print(paste("The left limit of the confidence interval", left)) #Ans: 8.54734549334714
print(paste("The right limit of the confidence interval", right)) #Ans: 12.2226545066529

So the confidence interval is = [8.54734549334714, 12.2226545066529]

** If the answer does not match please comment.