Statistics Experts?

[vt700guy]

I have a problem I'm hoping someone can help me with. The problem is laid out as follows:

 

The accompanying data are the times in seconds that it took a sample of employees to assemble a component at ABC industries. Assembly times are normally distributed. At the .05 significance level, can we conclude that the mean assembly time for this component is not equl to 3 minutes? Use Ha: U/= 180

(/= is my way of typing "not equal". Not sure how to do it otherwise)

 

The data points are:

190,199,198,176,180,174,181,183,208,188,198,165

 

I have already been provided the answer:

 

_

X=186.67

s=12.471

t=11,0.025 = 2.201 ( Straight from the t distribution table in the book)

 

Then of couse you plug those numbers into a formula and solve.

 

My question is this:

 

How do you calculate "S"?

For "t" why do you find the cross point with 11 and .025 in the table instead of 12?

 

Thanks in advance for anyone who has made it to this point and can help me. I have a very important test to pass that is likely to contain these types of problems.

[The Irish Doggy]

Aren't these answers in your text? / Can't you ask your prof?

 

Anyway, Looks like you are doing a t test and the concept of degrees of freedom is giving you trouble.

 

t = 11 is the degrees of freedom. you take n - 1 where n is the number of data points in your sample.

s is sample standard deviation of a sample taken from a population. You do it exactly the same as a normal standard deviation calculation but divide by 11 instead of 12 (again because it is a sample of a population you must use n-1).

 

ETA: You use 0.025 because it is half of 0.05. The table only gives you the value of / area beneath one tail of a normally distributed curve. Since you are looking for values beyond both the high and low tails with 0.05, you must divide the significance level by two to get the right answer from the table.

 

If you were looking for 0.1 significance level, you would read the table for t=11, 0.05

At 0.02 level, then t=11, 0.01

Edited September 15, 2009 by The Irish Doggy

[vt700guy]

Aren't these answers in your text? / Can't you ask your prof?

 

Anyway, Looks like you are doing a t test and the concept of degrees of freedom is giving you trouble.

 

t = 11 is the degrees of freedom. you take n - 1 where n is the number of data points in your sample.

s is sample standard deviation of a sample taken from a population. You do it exactly the same as a normal standard deviation calculation but divide by 11 instead of 12 (again because it is a sample of a population you must use n-1).

 

ETA: You use 0.025 because it is half of 0.05. The table only gives you the value of / area beneath one tail of a normally distributed curve. Since you are looking for values beyond both the high and low tails with 0.05, you must divide the significance level by two to get the right answer from the table.

 

If you were looking for 0.1 significance level, you would read the table for t=11, 0.05

At 0.02 level, then t=11, 0.01

 

This is not a class, there is not a professor per se'. It's a comprehensive final wrapping up a program. I do have access to the universities instructors but on a limited basis and I'd prefer to use them only when absolutely neccesary.

 

Thanks for the help on the problem. I'm starting to get it figured out. I have figured out the inputs on my financial calculator as well. Wish me luck, I have a couple weeks left before I take the test.

[The Irish Doggy]

Good luck.