Question

A brewery distributes beer in cans labeled 12 oz. The Bureau of Weights and Measures randomly selects 36 cans, measures their contents, and obtains a sample mean of 11.82 oz. Assume that σ is known to be 0.38 oz. At 0.01 significance level, can we conclude that the brewery is cheating consumers? Please show all 4 steps of the classical approach clearly.

Answer 1

Hypothesis

Null hypothesis:
`\mu = 12`

Alternative hypothesis:
`\mu \\eq 12`

Statistic

`t=(11.82-12)/((0.38)/(√(36)))=-2.84`

`df=n-1=36-1=35`

P value

The p value for this case would be given by:

`p_v =2*P(t_((35))<-2.84)=0.0075`

Conclusion

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 12 oz.

Explanation:

Information given

`\bar X=11.82` represent the sample mean

`s=0.38` represent the sample standard deviation

`n=36` sample size

`\mu_o =12` represent the value to verify

`\alpha=0.01` represent the significance level

t would represent the statistic

`p_v` represent the p value for the test

Hypothesis to verify

We want to cehck if the true mean is 12 or not, the system of hypothesis would be:

Null hypothesis:
`\mu = 12`

Alternative hypothesis:
`\mu \\eq 12`

Statistic

The statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing the info given we got:

`t=(11.82-12)/((0.38)/(√(36)))=-2.84`

The degrees of freedom are given by:

`df=n-1=36-1=35`

P value

The p value for this case would be given by:

`p_v =2*P(t_((35))<-2.84)=0.0075`

Conclusion

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 12 oz.