Random samples of size n are selected from a normal population whose standard deviation is know to be 2. Suppose you want 90% of the area under the sampling distribution of x bar to lie within + or - 1 unit of a population mean u. Find the minimum sample size n that satisfies the requirment.
To find the minimum sample size (n) that satisfies the requirement, we need to determine the appropriate value of the margin of error and then use the formula to calculate the sample size.
Given that we want 90% of the area under the sampling distribution of x-bar to lie within + or - 1 unit of the population mean (μ), the margin of error is 1 unit.
The formula to calculate the margin of error is:
Margin of Error = Critical Value * (Standard Deviation of the Population / Square Root of Sample Size)
Since we want 90% of the area under the sampling distribution to lie within the margin of error, the remaining 10% should be distributed evenly on both sides of the distribution.
To find the critical value corresponding to 90% confidence level (10% in each tail), we need to consult the Z-table (standard normal distribution table) or use a statistical software. The critical value corresponding to a 90% confidence level is approximately 1.645.
Now we can substitute the known values into the formula and solve for the sample size (n):
1 = 1.645 * (2 / √n)
To solve for n, we first square both sides of the equation:
1^2 = (1.645 * 2 / √n)^2
1 = 1.645^2 * (2 / √n)^2
1 = 2.705025 * (2^2 / n)
1 = 2.705025 * (4 / n)
Divide both sides of the equation by 2.705025:
1 / 2.705025 = 4 / n
0.3694 = 4 / n
Multiply both sides of the equation by n:
n * 0.3694 = 4
n = 4 / 0.3694
n ≈ 10.838
Since the sample size (n) must be a whole number, we round up to the nearest whole number:
n = 11
Therefore, the minimum sample size (n) that satisfies the requirement is 11.