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A pharmaceutical company manufactures a 200-milligram (mg) pain reliever. Company specifications require that the standard deviation of the amount of the active ingredient must not exceed 5 mg. The quality-control manager selects a random sample fo 30 tablets from a certain batch and finds that the sample standard deviation is 7.3 mg. Assume that the amount of the active ingredient is normally distributed. Determine whether the standard deviation of the amount of the active ingredient is greater than 5 mg at the 0.05 level of significance.

Respuesta :

Answer:

The standard deviation is not greater than 5 mg ([tex]\sigma = 2.1925[/tex])

Step-by-step explanation:

First we need to find the z-score for a 0.05 level of significance.

Looking in the z-table with alpha = 0.05, we have z = 1.645.

Now, finding the standard error of the mean, we have:

[tex]\sigma_{\bar{x}} = s_{x} / \sqrt{n}[/tex]

[tex]\sigma_{\bar{x}} = 7.3/ \sqrt{30}[/tex]

[tex]\sigma_{\bar{x}} = 1.3328[/tex]

The standard deviation with 0.05 level of significance would be:

[tex]\sigma = z*\sigma_{\bar{x}}[/tex]

[tex]\sigma = 1.645*1.3328[/tex]

[tex]\sigma = 2.1925[/tex]

This deviation is not greater than 5 mg.

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