Assume the population of wolves in a wilderness reserve area increases by the same number from month to month due to migration.
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For the given standard deviation and sample mean the two sided confidence interval on mean diameter foe 95% and 99% is given by ( 1.4983 , 1.5107 ) and ( 1.49635 , 1.51265 ) respectively.
As given in the question,
Confidence interval for population mean is given by:
± z × ( σ / √n)
Where,
σ = population standard deviation.
n= sample size
= Sample mean
Given values from the random sample size are :
σ = 0.01 inches
n= 10
= 1.5045
Critical value confidence interval (using x-value table)
z for 99% = 2.576 and Significance level = 0.01
z for 95% = 1.960 and Significance level = 0.05
99% two-sided confidence interval on the mean hole diameter is equal to :
± z × ( σ / √n)
= 1.5045 ± 2.576 × ( 0.01 / √10)
= 1.5045 ± 0.00815 (approximately)
= 1.5045 - 0.00815 , 1.5045 + 0.00815
= ( 1.49635 , 1.51265 )
95% two-sided confidence interval on the mean hole diameter is equal to :
± z × ( σ / √n)
= 1.5045 ± 1.960 × ( 0.01 / √10)
= 1.5045 ± 0.00620 (approximately)
= 1.5045 - 0.00620 , 1.5045 + 0.00620
= ( 1.4983 , 1.5107 )
Therefore, for the given standard deviation and sample mean the two sided confidence interval on mean diameter foe 95% and 99% is given by ( 1.4983 , 1.5107 ) and ( 1.49635 , 1.51265 ) respectively.
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