Respuesta :
Answer:
95% confidence interval for the population half-life based on this sample is [7.29 , 7.51].
Step-by-step 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. = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\bar X[/tex] = sample average half-life = 7.4 hours
[tex]\sigma[/tex] = population standard deviation = [tex]\sqrt{0.16}[/tex] = 0.4 hour
n = sample of people = 50
[tex]\mu[/tex] = 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, [tex]\mu[/tex] 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 < [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 1.96) = 0.95
P( [tex]-1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] <
P( [tex]\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] , [tex]\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ]
= [ [tex]7.4-1.96 \times {\frac{0.4}{\sqrt{50} } }[/tex] , [tex]7.4+1.96 \times {\frac{0.4}{\sqrt{50} } }[/tex] ]
= [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
