Answer to Question #241867 in Statistics and Probability for Md. Jahangir Alam

Question

A random sample of 25 tablets of buffered aspirin contains, on average, 325.05 mg

of aspirin per tablet, with a standard deviation of 55/10 mg. Find the 95% tolerance

limits that will contain 90% of the tablet contents for this brand of buffered aspirin.

Assume that the aspirin content is normally distributed.

Answer 1

We need to construct the $95\%$ confidence interval for the population mean $\mu.$

The critical value for $\alpha = 0.05$ and $df = n-1 = 24$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} =2.063899.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}},\bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(325.05-2.063899\times\dfrac{5.5}{\sqrt{25}},325.05+2.063899\times\dfrac{5.5}{\sqrt{25}})$$

$$=(322.780, 327.320)$$

Therefore, based on the data provided, the $95\%$ confidence interval for the population mean is $322.780<\mu< 327.320,$ which indicates that we are $95\%$ confident that the true population mean $\mu$ is contained by the interval $(322.780, 327.320).$