jaretktm2449
08.04.2020 •
Mathematics
Suppose we pick a simple random sample of 28 Major League Baseball players, and find that their average weight is 212.7 pounds, with a sample standard deviation of 15.6 pounds. Find a 99% confidence interval for the mean weight of all MLB players
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Ответ:
Step-by-step explanation:
We would use the t- distribution.
From the information given,
Mean, μ = 212.7 pounds
Standard deviation, σ = 15.6 pounds
number of sample, n = 28
Degree of freedom, (df) = 28 - 1 = 27
Alpha level,α = (1 - confidence level)/2
α = (1 - 0.99)/2 = 0.005
We will look at the t distribution table for values corresponding to (df) = 27 and α = 0.005
The corresponding z score is 2.77
We will apply the formula
Confidence interval
= mean ± z ×standard deviation/√n
It becomes
212.7 ± 2.77 × 15.6/√28
= 212.7 ± 2.77 × 2.948
= 212.7 ± 8.17
The lower end of the confidence interval is 212.7 - 8.17 = 204.53 pounds
The upper end of the confidence interval is 212.7 + 8.17 = 220.87 pounds
Ответ:
A reflects over the Y-axis, and B reflects over the x-axis.
Step-by-step explanation:
im a girl btw lol not a boy so sorry