Answer:
The answer is below
Step-by-step explanation:
Given that:
Mean (μ) = 3 ounces. standard deviation (σ) = 0.15, sample size (n) = 13 and confidence (C) = 98%
α = 1 - C = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01.
The z score of 0.01 (α/2) corresponds to the z score of 0.49 (0.5 - 0.01) which from the normal distribution table is 2.33.
The margin of error (E) is:
[tex]E=z_{0.01}*\frac{\sigma}{\sqrt{n} } = 2.33*\frac{0.15}{\sqrt{13} }=0.1\\[/tex]
The confidence interval = μ ± E = 3 ± 0.1 = (2.9, 3.1)
The confidence interval is between 2.9 ounce and 3.1 ounce
