If 6.2a=0.0143 then what is a?
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The value of mean, standard deviation and interval from the method of tree ring dating is 1273 AD, 37 years, [1250,1296].
According to the statement
we have to find that the standard deviation, mean and the intervals from the given data.
So, According to the given data from the method of tree ring dating
The value of mean is
x-bar = (1271 + 1208 + 1229 + 1299 + 1268 + 1316 + 1275 + 1317 + 1275) / 9 = x-bar = 1273 AD
And now we find standard deviation :
s = √∑(xi - x-bar) / (N - 1)
∑(xi - x-bar)^2 = (1271 - 1273)2 + (1208 - 1273)2 + (1229 - 1273)2 + ... + (1275 - 1273)2
∑(xi - x-bar)^2 = (-2)2 + (-65)2 + (-44)2 + ... + (2)2
∑(xi - x-bar)^2 = 4 + 4225 + 1936 + 676 + 25 + 1849 + 4 1936 + 4
∑(xi - x-bar)^2 = 10,659
Now,
s^2 = 10659/8 = 1332
s = 37 years
So, standard deviation is 37 years.
We need the t-distribution table since the standard deviation is unknown. Therefore, our degrees of freedom is 9 - 1 = 8 and the critical value is 1.860. Set up the confidence interval for the mean:
[x-bar ± t*(s/√n)] = [1273 ± 1.860*(37/√9)]
[x-bar ± t*(s/√n)] = [1250,1296]
So, The value of mean, standard deviation and interval from the method of tree ring dating is 1273 AD, 37 years, [1250,1296].
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Question:
The method of tree ring dating gave the following years A.D. for an archaeological excavation site. Assume that the population of x values has an approximately normal distribution.
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