What is the formula for the margin of error on mu for an exceptionally large sam

Question

What is the formula for the margin of error on mu for an exceptionally large sample? What is the difference between a parameter and an estimator? In your own words, define the Central Limit Theorem. Another study has been conducted regarding the weights of chickens in poultry farms in the U.S. A small family owned farm reported the following weights (lbs) of their 30 chickens. 3.93 3.99 3.96 3.78 3.96 3.94 3.88 3.96 4.02 3.93 4.03 4.06 3.84 4.10 3.94 3.75 4.02 4.09 4.98 4.06 4.17 3.84 3.92 4.12 3.75 3.99 4.01 4.20 3.72 3.88 a. Find a two sided 90% confidence interval of the mean weight of these chickens. b. Describe an appropriate interpretation of your computed confidence interval.

Explanation / Answer

Solution:

From the given weights
average of above mean = 3.960667

standard deviation = 0.121

a) value of Z for 90% CI = ±1.64485

hence Confidence interval = mean ± z*std dev/(n)^0.5
= 3.960667 ± 1.64485 * (0.121/30^0.5)
hence (LCL, UCL ) =(3.92433, 3.997004)

b) hence from above we can be certain upto 90% that mean weight of chicken population will be between 3.92433 and 3.997004

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