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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 :

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   [tex]\mu  = 0.86[/tex]

   The  sample is  mean is [tex]\=  x  =  0.8708[/tex]

  The standard deviation is  [tex]\sigma =  0.0068[/tex]

The  level of significance is  [tex]\alpha  =  0.01[/tex]

  The  sample size n =  3

The null hypothesis is  [tex]\mu =  0.86[/tex]

The alternative hypothesis  is [tex]\mu \ne  0.86[/tex]

Generally the test statistics is mathematically represented as  

    [tex]z =  \frac{\=  x  - \mu }{ \frac{\sigma}{\sqrt{n} } }[/tex]

=> [tex]z =  \frac{ 0.8708 - 0.86 }{ \frac{0.0068}{\sqrt{3} } }[/tex]

=>  [tex]z =  2.751  [/tex]  

Generally  p- value is mathematically represented as

       [tex]p-value =  2 P(z >2.751   )[/tex]

From the z-table  

        [tex]P(z >2.751 ) =  0.0029707[/tex]

So

     [tex]p-value =  2 *  0.0029707[/tex]

=>   [tex]p-value =  0.00594 [/tex]

From the obtained question we see that  [tex]p-value  <  \alpha[/tex]

Hence we reject the null hypothesis

 

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