# Specification Hours

• Dec 15th 2008, 10:46 PM
magentarita
Specification Hours
Random sample of 120 customers’s spent an average of 9.2 hours on their professional job with a sample standard deviation of 5.1 hours. Calculate the specification hours with confidence of 95%.
• Dec 17th 2008, 04:02 PM
meymathis
Quote:

Originally Posted by magentarita
Random sample of 120 customers’s spent an average of 9.2 hours on their professional job with a sample standard deviation of 5.1 hours. Calculate the specification hours with confidence of 95%.

I have not heard of specification hours. It looks like you are supposed to calculate the confidence interval with 95% confidence level.

The Central Limit Theorem (which the general rule of thumb says is ok to invoke for $N\geq 30$) tells us that $\frac{\bar{X}-\mu}{S \sqrt{N}}$ is approximately normally distributed with mean 0 and std dev 1.

So then the confidence interval is given by $\bar{X}\pm z_{\alpha/2}S/\sqrt{N}$ where $z_{\alpha/2}$ is defined by $\mathrm{P}(-z_{\alpha/2}\leq Z \leq z_{\alpha/2})=1-\alpha$ (Z is from the standard normal distribution). I'm assuming that you have already seen where these formulas came from.

So from a standard normal distribution table $z_{\alpha/2}=1.96$. So just plug away.
• Dec 21st 2008, 12:53 PM
magentarita
ok...
Quote:

Originally Posted by meymathis
I have not heard of specification hours. It looks like you are supposed to calculate the confidence interval with 95% confidence level.

The Central Limit Theorem (which the general rule of thumb says is ok to invoke for $N\geq 30$) tells us that $\frac{\bar{X}-\mu}{S \sqrt{N}}$ is approximately normally distributed with mean 0 and std dev 1.

So then the confidence interval is given by $\bar{X}\pm z_{\alpha/2}S/\sqrt{N}$ where $z_{\alpha/2}$ is defined by $\mathrm{P}(-z_{\alpha/2}\leq Z \leq z_{\alpha/2})=1-\alpha$ (Z is from the standard normal distribution). I'm assuming that you have already seen where these formulas came from.

So from a standard normal distribution table $z_{\alpha/2}=1.96$. So just plug away.

I thank you for your time and effort.