Question

A prescription allergy medicine is supposed to contain an average of 245 parts per million (ppm) of active ingredient. The manufacturer periodically collects data to determine if the production process is working properly. A random sample of 64 pills has a mean of 250 ppm with a standard deviation of 12 ppm.

Let µ denote the average amount of the active ingredient in pills of this allergy medicine. The null and alternative hypotheses are as H₀: µ = 245, H_a:µ ≠ 245. The level of significance is 1% .

The t-test statistic is 3.33 with a P-value of 0.0014. What is the correct conclusion?

		The mean amount of active ingredient in pills of this allergy medicine is equal to 245 ppm.
		The mean amount of active ingredient in pills of this allergy medicine is not equal to 245 ppm.
		The mean amount of active ingredient in pills of this allergy medicine is greater than 245 ppm.
		The mean amount of active ingredient in pills of this allergy medicine is equal to 250 ppm.

Answer 1

Solution :

Given that ,

= 245

= 250

= 12

n = 64

The null and alternative hypothesis is ,

H₀ :   = 245

H_a :    245

This is the two tailed test .

Test statistic = z

= ( - ) / / n

= ( 250 -245 ) / 12 / 64

= 3.33

The test statistic = 3.33

P - value = 2 * P(Z > 3.33 ) = 2 * 1 - P (Z < 3.33 )

= 2 * 1 - 0.9996

= 2 * 0.0004

= 0.0008

P-value = 0.0008

= 0.01

0.0008 < 0.01

P-value <

Reject the null hypothesis .

Conclusion : - There is sufficient evidence to test the claim .

The mean amount of active ingredient in pills of this allergy medicine is equal to 245 ppm.