Answer:
Following are the solution to the given question:
Explanation:
Performance of the student's t-test as
n<30
Calculating the Null hypothesis:
![\to H_0 : \mu =6](https://img.qammunity.org/2022/formulas/mathematics/high-school/p7jx7vcmatf7ysamv0pkj89xa3r9p94s8c.png)
Calculating the Alternative Hypothesis:
Calculating the level of significance
Calculating the test statistic:
![\to \bar{x}=(5.95 +6.10+ 5.98+ 6.01 +6.25+ 5.85 +5.91+ 6.05 +5.88+ 5.91)/(10)\\\\=(59.89)/(10)\\\\=5.989](https://img.qammunity.org/2022/formulas/mathematics/high-school/v291oayo86sdl36o542enq1bpixzwyyjb7.png)
Because of the population standard deviation, perform z test
Decision:
Comparison of p-value test statistics and decision-making.
Hypothesis P<0.05 Reject.
P>0.05 No hypothesis rejecting zero.
P-Value = 0.907653.
At p<0.05 the result is not significant.
Null hypothesis not to be rejected.
Accepting the null hypothesis.
Conclusion:
Its assertion that the containers were not refilled appropriately by the specified amount of 6 ounces/bottle doesn't contain substantial proof. Bottles with mean = 6 ounces are suitably filled.