Question

A local craft brewery claims the amount of beer in its bottles is 12oz (355ml). It knows that making false claims on its labels would result in serious penalties if it overstated the true volume. Every Monday morning, a sample of 30 bottles is taken to test the accuracy of their filling machines. Over the past few years of weekly sampling, they have calculated the standard deviation of the population to be σ = 1.6oz. They would like this to be a maximum level of variation in their filing process so, periodically, this is tested as well. This week, the sample resulted in a sample standard deviation of s = 2.

Answer 1

Answer:

a) The null and alternative hypothesis are:

$H_0:\sigma^2=\sigma_0^2\\\\H_a: \sigma^2>\sigma_0^2$

b) The statistic T is:

$T=\frac{n-1}{(s/\sigma)^2} =\frac{30-1}{(2.0/1.6)^2}=\frac{29}{1.25^2}=\frac{29}{1.5625}= 18.56$

c) The P-value for this test statistic T=18.59, and 29 degrees of freedom is P=0.93.

The P-value is bigger than the significant level, so the effect is not significant. The null hypothesis failed to be rejected.

d) The critical value for a one side test (α = 0.05, df=29) is:

$\chi^2_{29}=42.557$

As the T statistic is smaller than the critical value, it lies within the acceptance region. The null ypothesis is failed to be rejected.

The null hypothesis states that the sample standard deviation has no significant difference from the population standard deviation.

The alternative hypothesis states that in reality the sample standard deviation shows that the population standard deviation is not really 1.6 oz, but bigger.

e) The conclusion is that the sample result (standard deviation) is not significantly different from the populatiton standard deviation.

The accuracy of the machines don't show a significant change for this sample.

Step-by-step explanation:

The question is incomplete:

"Perform a test to determine if they are exceeding their maximum desired variance. Use α = 0.05.

a)Formulate the null and alternative hypothesis. Justify your formulation.

b)Calculate the test statistic.

c)Use the p-value approach to draw your conclusion.

d)Verify your conclusion using the critical value approach.

e)Interpret your conclusion."

The sample size is n=30.

The standard deviation of the population is σ=1.6 oz.

The sample standard deviation is s=2 oz.

We have to perform a one side (right-tail) test.

a) The null and alternative hypothesis are:

$H_0:\sigma^2=\sigma_0^2\\\\H_a: \sigma^2>\sigma_0^2$

The null hypothesis states that the sample standard deviation has no significant difference from the population standard deviation.

The alternative hypothesis states that in reality the sample standard deviation shows that the population standard deviation is not really 1.6 oz, but bigger.

b) The statistic T is:

$T=\frac{n-1}{(s/\sigma)^2} =\frac{30-1}{(2.0/1.6)^2}=\frac{29}{1.25^2}=\frac{29}{1.5625}= 18.56$

c) The P-value for this test statistic T=18.59, and 29 degrees of freedom is P=0.93.

The P-value is bigger than the significant level, so the effect is not significant. The null hypothesis failed to be rejected.

d) The critical value for a one side test (α = 0.05, df=29) is:

$\chi^2_{29}=42.557$

As the T statistic is smaller than the critical value, it lies within the acceptance region. The null ypothesis is failed to be rejected.

e) The conclusion is that the sample result (standard deviation) is not significantly different from the populatiton standard deviation.

The accuracy of the machines don't show a significant change for this sample.