A manufacturer of pharmaceutical products analyzes specimens from each batch of a product to verify the concentration of the active ingredient, required to be .80 grams per liter. Suppose the repeated measurements follow a Normal distribution. Fifteen specimens are taken from a batch and it is found that 0.82 grams per liter and s 0.048 grams per liter. Find a 99% confidence interval for μ, the true average concentration of the active ingredient.

So then we can conclude that at 99% of confidence the true mean is between (0.783;0.857) and the specification is satisfied since the value of 0.8 is on the confidence interval

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X= 0.82[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

s=0.048 represent the sample standard deviation

n=15 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=15-1=14[/tex]

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,14)".And we see that [tex]t_{\alpha/2}=2.977[/tex]

Now we have everything in order to replace into formula (1):

[tex]0.82-2.977\frac{0.048}{\sqrt{15}}=0.783[/tex]

[tex]0.82+2.977\frac{0.048}{\sqrt{15}}=0.857[/tex]

So on this case the 99% confidence interval would be given by (0.783;0.857)

So then we can conclude that at 99% of confidence the true mean is between (0.783;0.857) and the specification is satisfied since the value of 0.8 is on the confidence interval