# Clearly demonstrate 6stepsof Testing Hypotheses Regarding the Population Mean with o(standard deviation)known, using the classical or P-value approaA local juice manufacturer distributes juice in...

Clearly demonstrate 6stepsof Testing Hypotheses Regarding the Population Mean with o(standard deviation)known, using the classical or P-value approa

A local juice manufacturer distributes juice in bottles labeled 32 ounces. A government agency thinks that the company is cheating its customer. The agency selects 50 of these bottles, measures their contents, and obtains a sample mean of 32.6 ounces;the standard deviation, o, is 0.70 ounce. Use a 0.01 significance level to test the agency's claim that the compay is cheating its customers.

1. Null and alternative hypothese.

2. Level of significance

3. Test statistic

4. Critical Value (critical regions)

5 Decision

6. Conclusion

### 1 Answer | Add Yours

1. `H_0:mu=32` `H_1:mu<32` In words: the null hypothesis is that the average is 32oz, while the alternative hypothesis (and the claim in this case) is that the average is less than 32oz.

2. The level of significance is given as `alpha=.01`

3. Assuming that the population is approximately normal, and knowing the population standard deviation, we use the z-test:

The test statistic is `z=(32.6-32)/(.7/sqrt(50))~~6.06` (`z=(bar(x)-mu)/(sigma/sqrt(n))` )

4. The critical region is given by `z_(alpha/2)=z_(.005)=-2.58`

If you treat this as strictly a one-tailed test the critical region is `z_(.01)~~-2.33`

5. In either case, the test statistic is not in the critical region so we do not reject the null hypothesis.

6. There is not sufficient evidence to claim that the average is less than 32 oz.

** My calculator gives the test statistic at 6.060915267 and p=.9999999993 which is certainly not less than .01, so you would not reject the null hypothesis.

**Sources:**