Question

Antipsychotic drugs are widely prescribed for conditions such as schizophrenia and bipolar disease. An article reported on body composition and metabolic changes for individuals who had taken various antipsychotic drugs for short periods of time.

(a) The sample of 41 individuals who had taken aripiprazole had a mean change in total cholesterol (mg/dL) of 3.95, and the estimated standard error sD n was 3.478. Calculate a confidence interval with confidence level approximately 95% for the true average increase in total cholesterol under these circumstances. (Round your answers to two decimal places.) , mg/dL

(b) The article also reported that for a sample of 36 individuals who had taken quetiapine, the sample mean cholesterol level change and estimated standard error were 9.04 and 4.256, respectively. Make any necessary assumptions about the distribution of change in cholesterol level, test the hypothesis that the true average cholesterol level increases. State the appropriate hypotheses. (Use μD = μafter − μbefore.) H0: μD = 0 Ha: μD ≥ 0 H0: μD = 0 Ha: μD ≤ 0 H0: μD = 0 Ha: μD < 0 H0: μD = 0 Ha: μD > 0 H0: μD = 0 Ha: μD ≠ 0 Correct: Your answer is correct. Calculate the test statistic and P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.)

t =

P-value =

(c) For the sample of 45 individuals who had taken olanzapine, the article reported (7.630, 9.940) as a 95% CI for true average weight gain (kg). What is a 99% CI? (Round your answers to three decimal places.) , kg

Answer 1

a)

sample mean 'x̄=	3.950
sample size    n=	41.00
std deviation s=	3.478
std error ='sx=s/√n=	0.5432

for 95% CI; and 40 df, value of t=		2.0210
margin of error E=t*std error    =		1.098
lower bound=sample mean-E =		2.852
Upper bound=sample mean+E =		5.048
from above 95% confidence interval for population mean =(2.85,5.05)

b)

sample mean 'x̄=	9.040
sample size    n=	36.00
std deviation s=	4.256
std error ='sx=s/√n=	0.7093

test stat t ='(x-μ)*√n/sx=	12.74
p value      =	0.000

c)

sample mean =(9.940+7.630)/2=8.785

margin of error =(9.940-7.630)/2=1.155

for 95% CI; and 44 df, value of t=

2.0150

for 99% CI; and 44 df, value of t=

2.6920

margin of error for 99% CI =(2.692/2.015)*1.155 =1.543

99% CI =(8.785 -/+ 1.543) = (7.242 , 10.328)