CE 102What is the relationship between confidence intervals and t-distributions?

  • Thread starter bhoover05
  • Start date
  • Tags
    intervals
  • #1
bhoover05
20
0
I was under the impression that 95% C.I requires that the critical value in the error term comes from a t-distribution with 25 degrees of freedom.

I was taught in class today that a 25% CI requires that the critical value in the error term comes from a t-distribution with 23 degrees of freedom? I am unsure as to why this is?

~For example, Let's use this. . .
Average length of workweeks of 15 randomly selected employees in the mining industry and 10 randomly selected employees in the manufacturing industry were obtained.
Miners(x)=15 observation, Mean of 47.5, and Std. Dev. of 5.5
Manufacturers= 10 observations, Mean of 42.5, and Std. Dev of 4.9.
With this data I obtained that Sp 5.27, t=2.069, and a 95% CI =(0.55,9.45)

~I never used 23 degrees of freedom to obtain those answers, were are correct. Why should the 95% CI for the difference between x-y require the critical value in the error term comes from a t-distribtuion with 23 degrees of freedom?
 
Physics news on Phys.org
  • #2
I thought that Nu was directly related to n? Either n-1 or n-2 (can't remember which, statistics wasn't my favorite class).
 
  • #3
bhoover05 said:
I was under the impression that 95% C.I requires that the critical value in the error term comes from a t-distribution with 25 degrees of freedom.

I was taught in class today that a 25% CI requires that the critical value in the error term comes from a t-distribution with 23 degrees of freedom? I am unsure as to why this is?

~For example, Let's use this. . .
Average length of workweeks of 15 randomly selected employees in the mining industry and 10 randomly selected employees in the manufacturing industry were obtained.
Miners(x)=15 observation, Mean of 47.5, and Std. Dev. of 5.5
Manufacturers= 10 observations, Mean of 42.5, and Std. Dev of 4.9.
With this data I obtained that Sp 5.27, t=2.069, and a 95% CI =(0.55,9.45)

~I never used 23 degrees of freedom to obtain those answers, were are correct. Why should the 95% CI for the difference between x-y require the critical value in the error term comes from a t-distribtuion with 23 degrees of freedom?

The critical value depends on the number of degrees of freedom, which depends on the sample size. For a 95% CI (a t-distribution is used for small samples) the number of degrees of freedom will typically be [tex]n_1 + n_2 - 2[/tex], if I remeber correctly...this should be in your statistics book somewhere and you have to make some assumptions about the populations.

CS
 
  • #4
Oh that is so easy! I way over thought that. . . Thanks guys
 
  • #5
Ok. . . Now to expand from CI's to hypothesis testing. . .

I understand that for small n, and data approx. Normal, you use the formula
T= (x-bar - mu initial)/(s/ sqr n)

Now, what If i have no mu initial given. . .

Example-
Sample mean= 0.8
St. D= 0.1789
n=6
 
  • #6
"T= (x-bar - mu initial)/(s/ sqr n)

Now, what If i have no mu initial given. . .

Example-
Sample mean= 0.8
St. D= 0.1789
n=6"

In this case, I think what you want to do is _find_ mu initial. So, plug the values you have into the above formula. This gives,
T=(.8 - mu initial)/(.0730)

You also should have a size for the confidence level. Usually, it is 95% so I will use that.

Now, T is distributed as a t-distribution with _5_ degrees of freedom. (This is because don't know the _TRUE_ sigma. You then must estimate it by s, so you must subtract one from the number of data points in the sample to get the degrees of freedom.) In your table of t-distributions, find a number. That number, call it "talpha/2" has the following property.

The probability of T being larger than talpha/2 is .025.

(I think the above is an important phrase to remember.)


I looked up t in my table and found 2.571.

Then, 95% of the time,

-talpha/2<=T<=talpha/2

(The idea here is that the probability of being greater than talpha/2 is .025 and the probability of being less than -talpha/2 is also .025. Therefore, 95% of the time, T will be between -talpha/2 and talpha/2.)

This now says:

-2.571<=(.8-mu initial)/(.0730)<=2.571

Now solve this inequality for mu initial. You will then have a confidence interval for mu initial.

So, the procedure is:
1) Find the size of your confidence interval and subtract it from 1.
2) Divide that number by two.
3) Look up talpha/2 in a table with n-=1 degrees of freedom.
4) Next plug your other numbers into the formula (xbar - mu initial)/(s/sqr(n)). You will now have something only involving mu initial.
5) Put that formula between -talpha/2 and +talpha/2.
6) Solve for mu initial.

----------------------
Now, in your previous question, you asked why you needed 23 degrees of freedom rather than 24. That question looks like a t-test to me. You are trying to find out if there is a non-random difference between the length of work weeks of Manufacturers and of Miners.
In this case, there are two standard deviations you don't know. (You have an estimate s but you do not know the _TRUE_ standard deviation, sigma.) You must

(1)estimate them by their standard deviations and
(2)combine these estimates to get an estimate of the standard deviation of the distribution of the differences between the means.

Because you are using two estimates, you will need to subtract two degrees of freedom.

The rule-of-thumb is that, you must subtract one degree of freedom for each standard deviation you do not know for certain.
-------------------------

I hope this helps.
 
  • #7
bhoover05 said:
Ok. . . Now to expand from CI's to hypothesis testing. . .

I understand that for small n, and data approx. Normal, you use the formula
T= (x-bar - mu initial)/(s/ sqr n)

Now, what If i have no mu initial given. . .

Example-
Sample mean= 0.8
St. D= 0.1789
n=6

If [itex]\mu_0[/itex] is not given or assumed, then you have nothing to test. A hypothesis test is used to infer something about the population mean relative to [itex]\mu_0[/itex]. For example your null hypothesis may be that [itex]\mu = 0[/itex] and your alternative hypothesis [itex]\mu > 0[/itex]. If your test statistic (T) falls in the rejection region, then you can infer that the population mean, [itex]\mu[/itex], is greater than 0. Note that we tested the hypothesis that the population mean was equal to 0 (i.e. you selected the value for [itex]\mu_0[/itex] which was 0 in this example).

Alternatively, [itex]\mu_0[/itex] can be any value you wish to test the population mean against.

Refer to page 6 of this http://www.sjsu.edu/faculty/gerstman/StatPrimer/hyp-test.pdf" for more info.

Hope that helps.

CS
 
Last edited by a moderator:

1. What is a 95% Confidence Interval?

A 95% confidence interval is a range of values that is likely to contain the true value of a population parameter with a confidence level of 95%. This means that if the same population is sampled multiple times and a confidence interval is calculated each time, 95% of those intervals will contain the true population parameter.

2. How is a 95% Confidence Interval calculated?

A 95% confidence interval is calculated using a formula that takes into account the sample size, sample mean, and standard deviation of the sample. This formula is based on the Central Limit Theorem, which states that the sampling distribution of the mean will be approximately normal for a large sample size.

3. Why is a 95% Confidence Interval used?

A 95% confidence interval is commonly used because it provides a balance between precision and reliability. It is a narrow enough range to give a precise estimate of the population parameter, but also has a high enough confidence level to ensure that the interval is likely to contain the true value.

4. What does a 95% Confidence Interval tell us?

A 95% confidence interval tells us that we are 95% confident that the true population parameter falls within the calculated range. It also gives us an idea of the precision of our estimate, as a narrower interval indicates a more precise estimate.

5. How can a 95% Confidence Interval be interpreted?

A 95% confidence interval can be interpreted as a range of values that we are 95% confident contains the true population parameter. It is important to note that the true value may or may not be within the interval, but the probability of it being within the interval is 95%. Additionally, a wider interval may indicate a larger margin of error in the estimate of the population parameter.

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
Replies
2
Views
4K
Replies
1
Views
2K
Replies
1
Views
952
  • Set Theory, Logic, Probability, Statistics
Replies
9
Views
3K
  • Precalculus Mathematics Homework Help
Replies
1
Views
2K
  • Precalculus Mathematics Homework Help
2
Replies
37
Views
6K
  • Art, Music, History, and Linguistics
Replies
1
Views
1K
  • Poll
  • Science and Math Textbooks
Replies
1
Views
2K
Back
Top