A supervisor of White Industries is considering a new method of
assembling its cart product.  The present method requires 42.3
minutes, on the average, to assemble a cart. The mean assembly time
for a random sample of 24 carts, using the new method, was 40.6
minutes, with a standard deviation of 2.7 minutes.  Using the 0.05
level of significance, can we conclude that the assembly time using
the new method is faster?

---

Request for Question Clarification by scriptor-ga on 25 Aug 2005 15:59 PDT

Google Answers discourages and may remove questions that are homework
or exam assignments.

Scriptor

Lola5 ?

What we?re seeking here is the confidence interval with p = 0.05. 
Restated, it?s a 95% confidence that the same fall within the range. 
It's calculated as:

mean +/- T-value * SD

BMJ.com
?Statements of probability and confidence intervals? (undated)
http://bmj.bmjjournals.com/collections/statsbk/4.shtml


The small sample size here calls for the use of the Student?s
T-distribution, which narrows the range of values depending on the
sample size.  It's important to remember that the degrees of freedom
are one less than the sample size:

Statsoft Electronic Textbook
?Student?s T Table? (2003)
http://www.statsoft.com/textbook/sttable.html

Thus at p < 0.05, the mean is 40.6 +/-  (1.713872) * 2.7 

So, anything in that range is statistically expected and we can?t
reject the hypothesis H0 (also called the null hypothesis) ? that this
new assembly technique is the same as the 42.3 minutes of the old
technique.  Only those values outside the range would support
hypothesis H1 ?  the alternate hypothesis --

Statistics Glossary
?Hypothesis Testing?
http://www.cas.lancs.ac.uk/glossary_v1.1/hyptest.html

So, values between 40.6 +/- 4.63 minutes or in the range of 35.97 ?
45.23 are indistinguishable from our sample.  Of course 42.3 minutes
is well within that range, so we can?t conclude that we?re doing any
better.


Google search strategy:
Probablility + ?confidence interval?
T distribution

Best regards,

Omnivorous-GA