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Sagot :
Answer:
The calculated value Z = 11.792 >1.96 at a 0.05 level of significance.
The null hypothesis is rejected
The alternative hypothesis is accepted at a 0.05 level of significance.
The mean weight is less than 16 ounces
Step-by-step explanation:
Step:-1
Given that the mean of the Population = 16
Given that the standard deviation of the process
σ = 0.03 ounce
Given that the mean of the sample = 16.05 ounces
Null hypothesis: H₀:μ > 16 ounces
Alternative HypothesisH₁: μ <16 ounces
Step:-2
Test statistic
[tex]Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }[/tex]
[tex]Z = \frac{16.05 -16}{\frac{0.03}{\sqrt{50} } }[/tex]
Z = 11.792
Level of significance = 0.05
Z₀.₀₅ = 1.96
Final answer:-
The calculated value Z = 11.792 >1.96 at a 0.05 level of significance.
The null hypothesis is rejected
The alternative hypothesis is accepted at a 0.05 level of significance.
The mean weight is less than 16 ounces
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