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 90 mg. It is known that the standard deviation in the amount of the chemical is 6 mg. A random sample of 31 batches of the new drug is tested and found to have a sample mean concentration of 92.2 mg of the active chemical.
Calculate the 95% confidence interval for the mean amount of the active chemical in the drug.

Answer 1

Answer:

95% confidence interval for the mean amount of the active chemical in the drug = [ 90.088 , 94.312 ]

Step-by-step explanation:

We are given that a lab is testing the amount of a certain active chemical compound in a particular drug that has been recently developed. It is known that the standard deviation in the amount of the chemical is 6 mg.

A random sample of 31 batches of the new drug is tested and found to have a sample mean concentration of 92.2 mg of the active chemical i.e.;

Population standard deviation, $\sigma$ = 6 mg

Sample mean, $Xbar$ = 92.2 mg

Sample size, n = 31

Now, the pivotal quantity for 95% confidence interval is given by;

          $\frac{Xbar -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

So, 95% confidence interval for the mean amount of the active chemical in the drug is given by;

P(-1.96 < N(0,1) < 1.96) = 0.95

P(-1.96 < $\frac{Xbar -\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P(-1.96 * $\frac{\sigma}{\sqrt{n} }$ < $Xbar - \mu$ < 1.96 * $\frac{\sigma}{\sqrt{n} }$ ) = 0.95

P(Xbar - 1.96 * $\frac{\sigma}{\sqrt{n} }$ < $\mu$ < Xbar + 1.96 * $\frac{\sigma}{\sqrt{n} }$ ) = 0.95

95% confidence interval for $\mu$ = [Xbar - 1.96 * $\frac{\sigma}{\sqrt{n} }$ , Xbar + 1.96 * $\frac{\sigma}{\sqrt{n} }$ ]

                                                  = [ 92.2 - 1.96 * $\frac{6}{\sqrt{31} }$ , 92.2 + 1.96 * $\frac{6}{\sqrt{31} }$ ]

                                                  = [ 90.088 , 94.312 ]