Answer:
Margin of error = 2.74 ounces
Step-by-step explanation:
We are given the following in the question:
Sample mean, [tex]\bar{x}[/tex] = 73 ounces
Sample size, n = 32
Alpha, α = 0.05
Population standard deviation, σ = 7.9 ounces
Margin of error:
[tex]z_{critical}\times \dfrac{\sigma}{\sqrt{n}}[/tex]
[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]
Putting the values, we get,
[tex]1.96\times (\dfrac{7.9}{\sqrt{32}} ) =2.7372 \approx 2.74[/tex]
Thus, the maximum margin of error associated with a 95% confidence interval for the true population mean turtle weight is 2.74 ounces
The maximal margin of error associated with a 95% confidence interval is 2.74
Given that n(sample size) = 32, standard deviation (σ) = 7.9
Confidence (C) = 95% = 0.95
α = 1 - C = 0.05
α/2 = 0.05/2 = 0.025
The z score of α/2 corresponds to the z score of 0.475 (0.5 - 0.025) which is equal to 1.96
The margin of error (E) is:
[tex]E=z_\frac{\alpha}{2}*\frac{\sigma}{\sqrt{n} } \\\\E=1.96*\frac{7.9}{\sqrt{32} } =2.74[/tex]
Therefore the maximal margin of error associated with a 95% confidence interval is 2.74
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