Question

Assume that a sample is used to estimate a population proportion μ . Find the margin of error M.E. that corresponds to a sample of size 722 with a mean of 54.2 and a standard deviation of 13.1 at a confidence level of 90%.

Answer 1

Answer:

$MoE = 1.645\cdot \frac{13.1}{\sqrt{772} } \\\\MoE = 1.645\cdot 0.47147\\\\MoE = 0.776$

Step-by-step explanation:

Since the sample size is quite large, we can use the z-distribution.

The margin of error is  given by

$MoE = z_{\alpha/2}(\frac{s}{\sqrt{n} } )$

Where n is the sample size, s is the sample standard deviation and $z_{\alpha/2}$ is the z-score corresponding to a 90% confidence level.

The z-score corresponding to a 90% confidence level is

Significance level = α = 1 - 0.90= 0.10/2 = 0.05

From the z-table at α = 0.05

z-score = 1.645

$MoE = 1.645\cdot \frac{13.1}{\sqrt{772} } \\\\MoE = 1.645\cdot 0.47147\\\\MoE = 0.776$

Therefore, the margin of error is 0.776.