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Sagot :
Answer:
Following are the solution to the given question:
Step-by-step explanation:
Performance of the student's t-test as [tex]n=10[/tex]
n<30
Calculating the Null hypothesis:
[tex]\to H_0 : \mu =6[/tex]
Calculating the Alternative Hypothesis:
[tex]\to H_1 : \mu \neq 6[/tex]
Calculating the level of significance [tex]=0.05[/tex]
Calculating the test statistic:
[tex]\to \bar{x}=\frac{5.95 +6.10+ 5.98+ 6.01 +6.25+ 5.85 +5.91+ 6.05 +5.88+ 5.91}{10}\\\\=\frac{59.89}{10}\\\\=5.989[/tex]
Because of the population standard deviation, perform z test
[tex]z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]=\frac{5.989-6}{\frac{0.3}{\sqrt{10} }} \\\\=-0.116[/tex]
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.
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