Question

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.  

Answer 1

Number of tablets in the random sample, n = 30

For testing of the above problem,

The null hypothesis would be, H0: ≤ 5 mg against the

Alternative hypothesis, Ha: > 5 mg

The sample standard deviation, s’ = 7.3 mg

The test statistic for the above problem, c = [(n - 1)(s’^2)] / (^2) which under the null hypothesis follows chi – square distribution with (n - 1) degrees of freedom

The value of the test statistic, c = [(30 - 1)(7.3^2)]/(5^2) = 61.816

The critical value for the problem, 0.05, 29 = 42.557 (value taken from chi – square distribution table)

If the value of the test statistic is greater than the critical one would reject the null hypothesis else one fails to reject the null hypothesis.

Since, the value of the test statistic > the critical value

The null hypothesis is rejected.

There is evidence that the standard deviation of the amount of active ingredient is not ≤ 5 mg, at 5% level of significance. So, the standard deviation might be > 5 mg.