Question

A friend of yours was interested in computing a confidence interval for the mean student debt accumulated by NC State graduates during their undergraduate education. They collected a random sample of 40 students, and asked each how much student debt they had; all responded and responded honestly. A histogram of the data indicated it was normally distributed. They computed a 95% confidence interval of ($36,400, $41,650). However, they were unsatisfied with the margin of error, wanting a more precise interval. Referring to their ST 311 notes, they noted that in order to reduce the margin of error by half, they needed to increase their sample size by a factor of 4. So, they collected another random sample of 160 students and found another 95% confidence interval, which was much narrower, having a margin of error of $1,318. They are satisfied with the precision of the confidence interval, but are a little confused : it’s not quite half the original margin of error. Explain why the margin of error wasn’t quite reduced by a factor of 0.5. Be specific.

Answer 1

Explained below.

Explanation:

The (1 - α)% confidence interval for the population mean is:

`CI=\bar x\pm z_(\alpha/2)*(\sigma)/(√(n))`

The margin of error for this interval is:

`MOE= z_(\alpha/2)*(\sigma)/(√(n))`

The factors affecting the MOE are:

• The sample size (n) is inversely proportional to the MOE, thus increasing n decreases the MOE and vice-versa.
• Variance (σ²) is the square of S.D. So if we increase σ² then σ increases. Since σ is directly proportional to MOE, increasing σ increases MOE.
• The confidence level is the value of (1 – α). If we increase the confidence level then the critical value increases and thus MOE also increases.

Since the population standard deviation is not provided it is obvious the sample standard deviation is used to estimate the value of σ.

On increasing the value of σ the MOE will also increase.

As the sample size is inversely proportional, increasing the sample size by a factor of 4 will reduce the margin of error by half. But since σ is also increasing there are two factors affecting the MOE together.

Therefore reducing the MOE by more than half the original value.