Question

The process of producing pain-reliever tablets yields tablets with varying amounts of the active ingredient. The manufacturer claims each tablet has at least 200 milligrams of the active ingredient. The consumer Watchdog Bureau assumes the manufacturer claim is correct, but occasionally tests samples of the tablets to ensure they contain enough of the ingredient. The Consumer Watchdog Bureau tests a random sample of 70 tablets. The sample mean content of the active ingredient is 205.7 milligrams, while the sample standard deviation is 21 milligrams. What is the p-value for this test?

Answer 1

Answer:

The  p-value is  $p-value = 0.013167$

Step-by-step explanation:

From the question we are told that

     The  population mean is $\mu$ =  200 milligrams

     The sample size is $n = 70$

     The sample  mean is  $\= x = 205.7$

      The  sample standard deviation is  $\sigma = 21 \ milligram$

Generally the Null hypothesis is  mathematically represented as

      $H_o : \mu = 200$

The Alternative  hypothesis is  

     $H_a : \mu < 200$

The test statistics is mathematically represented as

       $t_s = \frac{\= x - \mu }{\frac{\sigma}{\sqrt{n} } }$

substituting values  

     $t_s = \frac{ 205.7 - 200 }{\frac{21}{\sqrt{70} } }$

     $t_s = 2.270$

Now the p-value is mathematically represented as

      $p-value = P(Z \le t_s )$

substituting values  

      $p-value = P(Z \le 2.270 )$

Using the Excel function[=NORMDIST(2.270)] to calculate the p-value we obtain that

       $p-value = 0.013167$