Amanufacturer 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 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

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".

represent the sample mean

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:

  (1)

In order to calculate the critical value we need to find first the degrees of freedom, given by:

Since the Confidence is 0.99 or 99%, the value of and , 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

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

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

tube or goog would be a great resource to with that question