Question

1. referring to question 3, what happened to the standard error when the sample size (n) was increased? use the formula to justify why this happened. standard deviation decreased

Explanation:

Step1: Recall Standard Error Formula

The formula for the standard error (SE) of the mean is \( SE=\frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation and \( n \) is the sample size.

Step2: Analyze the Effect of Increasing \( n \)

As \( n \) (the sample size) increases, the denominator \( \sqrt{n} \) in the formula \( SE = \frac{\sigma}{\sqrt{n}} \) becomes larger. Since \( \sigma \) (assuming it remains constant) is divided by a larger number, the value of \( SE \) will decrease. For example, if \( \sigma = 10 \) and \( n = 4 \), then \( SE=\frac{10}{\sqrt{4}} = 5 \). If \( n \) increases to \( 16 \), then \( SE=\frac{10}{\sqrt{16}} = 2.5 \), which is smaller than 5. So as sample size \( n \) increases, the standard error decreases because the formula shows that standard error is inversely related to the square root of the sample size.

Answer:

When the sample size \( n \) is increased, the standard error decreases. This is justified by the standard error formula \( SE=\frac{\sigma}{\sqrt{n}} \): as \( n \) (the sample size) increases, the denominator \( \sqrt{n} \) increases (for a constant population standard deviation \( \sigma \)), so the overall value of \( SE \) (the quotient \( \frac{\sigma}{\sqrt{n}} \)) decreases.