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

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  <  \alpha`

Hence we reject the null hypothesis