The pharmacy care study reported by Lee et al., 2006 reported medication adheren

Question

The pharmacy care study reported by Lee et al., 2006 reported medication adherence information (% adherent) on older adults who were taking multiple medications. The following confidence interval was obtained based on data from their study.

Baseline medication adherence at completion of run-in phase (n=179), mean (SD): 61.2 (13.5)

Which answer below is most accurate?

This confidence interval estimates the population proportion for a single sample.

We are confident that 95% of the participants had adherence between 59.2 and 63.2

We are 95% confident that the mean adherence level is between 59.2 and 63.2 for the population.

None of the above

Adherence was likely to be greater than 61.2%, for 95% of this sample.

This confidence interval estimates the population proportion for a single sample.

We are confident that 95% of the participants had adherence between 59.2 and 63.2

We are 95% confident that the mean adherence level is between 59.2 and 63.2 for the population.

None of the above

Adherence was likely to be greater than 61.2%, for 95% of this sample.

Explanation / Answer

TRADITIONAL METHOD
given that,
sample mean, x =61.2
standard deviation, s =13.5
sample size, n =179
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 13.5/ sqrt ( 179) )
= 1.009
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 178 d.f is 1.973
margin of error = 1.973 * 1.009
= 1.991
III.
CI = x ± margin of error
confidence interval = [ 61.2 ± 1.991 ]
= [ 59.209 , 63.2 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =61.2
standard deviation, s =13.5
sample size, n =179
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 178 d.f is 1.973
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 61.2 ± t a/2 ( 13.5/ Sqrt ( 179) ]
= [ 61.2-(1.973 * 1.009) , 61.2+(1.973 * 1.009) ]
= [ 59.209 , 63.2 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 59.209 , 63.2 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean

We are 95% confident that the mean adherence level is between 59.2 and 63.2 for the population.

We are 95% confident that the mean adherence level is between 59.2 and 63.2 for the population.

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