Question

An analytical laboratory is asked to evaluate the claim that the concentration of the active ingredient in a specimen is 86%. The lab makes 3 repeated analyses of the specimen. The sample mean result is 0.8708. The true concentration is the mean m of the population of all analyses of the specimen. The standard deviation of the analysis process is known to be LaTeX: \sigmaσ = 0.0068. Is there significant evidence at the 1% level that LaTeX: \muμ is different than 0.86? (assume normality)

Answer 1

Answer:

Decision rule

   Reject null hypothesis

Conclusion

There is significance evidence that the true mean is different from 0.86

Step-by-step explanation:

From the question we are told that

  The population mean   $\mu = 0.86$

   The  sample is  mean is $\= x = 0.8708$

  The standard deviation is  $\sigma = 0.0068$

The  level of significance is  $\alpha = 0.01$

  The  sample size n =  3

The null hypothesis is  $\mu = 0.86$

The alternative hypothesis  is $\mu \ne 0.86$

Generally the test statistics is mathematically represented as  

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

=> $z = \frac{ 0.8708 - 0.86 }{ \frac{0.0068}{\sqrt{3} } }$

=>  $z = 2.751$  

Generally  p- value is mathematically represented as

       $p-value = 2 P(z >2.751 )$

From the z-table  

        $P(z >2.751 ) = 0.0029707$

So

     $p-value = 2 * 0.0029707$

=>   $p-value = 0.00594$

From the obtained question we see that  $p-value < \alpha$

Hence we reject the null hypothesis