Suppose a team of researchers is studying the half-life of a drug in the human body (i.e. how long it takes for 1 2 of the drug to be broken down by the body). They take 50 peop…

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asked May 28, 2021 55.8k views

1 Answer

Sam Makin · Answer 1 · 2021-05-31T18:28:27+0000

Answer:

95% confidence interval for the population half-life based on this sample is [7.29 , 7.51].

Explanation:

We are given that the average half-life to be 7.4 hours. Suppose the variance of half-life is known to be 0.16.

They take 50 people, administer a standard dose of the drug, and measure the half-life for each of these people.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. =
$(\bar X-\mu)/((\sigma)/(√(n) ) )$ ~ N(0,1)

where,
$\bar X$ = sample average half-life = 7.4 hours

$\sigma$ = population standard deviation =
√(0.16) = 0.4 hour

n = sample of people = 50

$\mu$ = population mean

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean,
$\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 <
$(\bar X-\mu)/((\sigma)/(√(n) ) )$ < 1.96) = 0.95

P(
$-1.96 * {(\sigma)/(√(n) ) }$ <
${\bar X-\mu}$ <

P(
$\bar X-1.96 * {(\sigma)/(√(n) ) }$ <
$\mu$ <
$\bar X+1.96 * {(\sigma)/(√(n) ) }$ ) = 0.95

95% confidence interval for
$\mu$ = [
$\bar X-1.96 * {(\sigma)/(√(n) ) }$ ,
$\bar X+1.96 * {(\sigma)/(√(n) ) }$ ]

= [
7.4-1.96 * {(0.4)/(√(50) ) } ,
7.4+1.96 * {(0.4)/(√(50) ) } ]

= [7.29 , 7.51]

Therefore, 95% confidence interval for the population half-life based on this sample is [7.29 , 7.51].

The length of this interval is = 7.51 - 7.29 = 0.22