Question 1204280: You intend to estimate a population mean
μ with the following sample.
24.7
42.1
36
24.3
-7.8
32.2
40.4
You believe the population is normally distributed. Find the 80% confidence interval. Enter your answer as an open-interval. This means enter your answer inside of parentheses accurate to two decimal places (because the sample data are reported accurate to one decimal place).
For example if the lower limit is 12.23 and the upper limit is 15.34 enter your answer as (12.32,15.34)
80% C.I. =
Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places. Did you enter your answer as an open-interval as described above?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! mean is equal to 27.414285714286
standard deviatiion is equal to 17.019736442478
sample size is 7.
t-score 80% confidence interval with 6 degrees of freedom = plus or minus t = 1.439755738.
i rounded intermediate values to 6 decimal places.
that's more than enough detail for a final rounding of 2 or 3 decimal digits.
the rounded values used are:
mean = 27.414286
standard deviation = 17.019736
degrees of freedom = sample size of 7 minus 1 = 6
critical t-score = plus oe minus 1.439756
t-score formula used is t = (x - m) / s
t is the t-score
x is the raw score
m is the mean
s is the standard error
to solve for x, the formula bgecomes:
x = s * t + m
standard error = standard deviation / sqrt(sample size).
that is equal to 17.019736 / sqrt(7) = 6.432856
on the high side of the confidence interval, you get x = s * t + m becomes:
x = 6.432856 * 1.439756 + 27.414286 = 36.676029.
on the low side of the confidence interval, you get x = s * t + m becomes:
x = 6.432856 * -1.439756 + 27.414286 = 18.152543.
your solution is 80% confidence interval for the sample mean = (18.15,36.68).
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