Question

We would like to estimate the true mean weight of 20,000 ostriches. Earlier data indicates that the ostrich weights follow Normal distribution a standard deviation of 7 kg. From a specialized farm, a group of 196 ostriches were selected randomly to measure the precise weights, X₁,…,X₂₀₀;. You have found that, μ̂₁=65.03 kg. Assuming the confidence level 95% around the true mean (expected value) of the ostriche' weights, select the correct closest confidence interval from the following:

	1.	[56.06, 74.00]
	2.	[64.05, 66.01]
	3.	[62.02, 68.04]
	4.	[60.09, 69.97]
	5.	[59.05, 71.01]

Answer 1

As per the provided information,

Sample mean, is 65.03 kg

Sample size, n is 196

Population standard deviation,   is 7 kg

Confidence level (c) is 0.95

The significance level,

At the significance level 0.05, the two tailed critical value obtained from z-table is 1.96.

The 95% confidence interval for the true mean (expected value) of the ostriche' weights,

Thus, the 95% confidence interval for the true mean (expected value) of the ostriche' weights is (64.08 kg, 66.01 kg).

Therefore, the closest confidence interval is [64.05, 66.01].

Hence the option 2 is correct.