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Sagot :
There is a 95% chance that the mean length of all term papers lies between confidence limits (11.58,17.82) for the given data:
14, 20, 25, 10, 16, 8, 15, 12, 18, 9
We assume the length of paper has an approximately normal distribution.
Sample size = 10
The sample mean for the given data = 14.7
The standard deviation for the given data = 5.04
Critical value is obtained from standard normal distribution at the level of significance 0.05.
The critical value is 1.96.
The 95% confidence interval for the mean length of all term papers is calculated as follows:
P(14.7 - (1.96 x (5.04/√10)) < µ < 14.7 + (1.96 x (5.04/√10))) = 0.95
P(11.58 < µ < 17.82) = 0.95
To know more about Critical value here:
https://brainly.com/question/12069922
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