A random sample of 150 students has a test score average with a standard deviation of 9.9. Find the margin of error of c=0.95

Question

stat test hard and don't get it 

Answer 1

Hi Jordan,

 

Statistical concepts are sometimes difficult to make sense of. Before going into formulas, I will try to describe margin of error in lay terms. You can think of it as an indicator which represents the likelihood (i.e., not certainty) that the result obtained by sampling a population is close to the result that would have been obtained if we had conducted a census (i.e. surveying the entire population).

 

And, we can obtain this likelihood at different confidences like 99%, 95%, 90% and so on. In this question we are asked for 95% confidence. According to normal distribution, there is a z- value associated with every confidence level. For 95% confidence level, z value is 1.96.

 

Margin of error = Critical value x Standard error of the statistic

 

Remember from stats again that standard error, SE = s / sqrt(n) where s is standard deviation and n is sample size

SE = 9.9/sqrt(150)

SE = 0.808

then margin of error = 1.96*0.808 = 1.58

 

which means that we can be 95% confident that the mean test score in the population is plus or minus 1.58 of the average obtained from the sampling of these 150 students. 

 

Hope this makes sense!