Question

A lab is testing the amount of a certain active chemical compound in a particular drug that has been recently developed. The manufacturer claims that the average amount of the chemical is 95 mg. It is known that the standard deviation in the amount of the chemical is 5 mg. A random sample of 25 batches of the new drug is tested and found to have a sample mean concentration of 89.9 mg of the active chemical.
Calculate the 95% confidence interval for the mean amount of the active chemical in the drug. Give your answers to 2 decimal places. (suppose to be number less than and equal to mean less than and equal to another number)

Answer 1

Answer:

The 95% confidence interval is "87.94, 91.86".

Step-by-step explanation:

The given values are:

Sample batches,

n = 25

Sample mean concentration,

$\bar{x}$ = 89.9 mg

Amount of chemical,

σ = 5

α = 0.05

The critical value from Z table will be:

=  $Z(\frac{\alpha}{2} )$

=  $Z(\frac{0.05}{2} )$

=  $1.96$

Now,

The confidence interval will be:

=  $\bar{x} \pm Z(\frac{\alpha}{2} )\times \frac{\sigma}{\sqrt{n} }$

On substituting the values, we get

=  $89.9 \pm Z(\frac{0.05}{2})\times \frac{5}{\sqrt{25}}$

=  $89.9 \pm Z(\frac{0.05}{2})\times \frac{5}{5}$

=  $89.9 \pm1.96\times 1$

Lower limit will be:

=  $89.9 -1.96\times 1$

=  $87.94$

Upper limit will be:

=  $89.9 +1.96\times 1$

=  $91.86$