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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)

User Mezoni
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1 Answer

5 votes

Answer:

Decision rule

Reject null hypothesis

Conclusion

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

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 \\e  0.86

Generally the test statistics is mathematically represented as


z =  (\=  x  - \mu )/( (\sigma)/(√(n) ) )

=>
z =  ( 0.8708 - 0.86 )/( (0.0068)/(√(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 &nbsp;< &nbsp;\alpha

Hence we reject the null hypothesis

User Tomblue
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