The method of tree ring dating gave the following years A.D. for an archaeologic

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. 1194 1292 1187 1229 1268 1316 1275 1317 1275

(a) Use a calculator with mean and standard deviation keys to find the sample mean year x and sample standard deviation s. (Round your answers to the nearest whole number.)

x =       A.D.

s =       yr

(b) Find a 90% confidence interval for the mean of all tree ring dates from this archaeological site. (Round your answers to the nearest whole number.)

lower limit            A.D.

upper limit           A.D.

Explanation / Answer

PART A.
Mean = Sum of observations/ Count of observations
Mean = (1194 + 1292 + 1187 + 1229 + 1268 + 1316 + 1275 + 1317 + 1275 / 9) = 1261.4444
Variance
Step 1: Add them up
1194 + 1292 + 1187 + 1229 + 1268 + 1316 + 1275 + 1317 + 1275 = 11353
Step 2: Square your answer
11353*11353 =128890609
…and divide by the number of items. We have 9 items , 128890609/9 = 14321178.7778
Set this number aside for a moment.
Step 3: Take your set of original numbers from Step 1, and square them individually this time
1194^2 + 1292^2 + 1187^2 + 1229^2 + 1268^2 + 1316^2 + 1275^2 + 1317^2 + 1275^2 = 14339729
Step 4: Subtract the amount in Step 2 from the amount in Step 3
14339729 - 14321178.7778 = 18550.2222
Step 5: Subtract 1 from the number of items in your data set, 9 - 1 = 8
Step 6: Divide the number in Step 4 by the number in Step 5. This gives you the variance
18550.2222 / 8 = 2318.7778
Step 7: Take the square root of your answer from Step 6. This gives you the standard deviation
48.1537
PART B.
TRADITIONAL METHOD
given that,
sample mean, x =1261.4444
standard deviation, s =48.1537
sample size, n =9
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 48.1537/ sqrt ( 9) )
= 16.051
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 8 d.f is 1.86
margin of error = 1.86 * 16.051
= 29.855
III.
CI = x ± margin of error
confidence interval = [ 1261.4444 ± 29.855 ]
= [ 1231.589 , 1291.3 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =1261.4444
standard deviation, s =48.1537
sample size, n =9
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 8 d.f is 1.86
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 1261.4444 ± Z a/2 ( 48.1537/ Sqrt ( 9) ]
= [ 1261.4444-(1.86 * 16.051) , 1261.4444+(1.86 * 16.051) ]
= [ 1231.589 , 1291.3 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 1231.589 , 1291.3 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean

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