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95% Confidence Intervals

  1. Dec 12, 2008 #1
    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, Lets 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?
  2. jcsd
  3. Dec 12, 2008 #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).
  4. Dec 12, 2008 #3


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    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.

  5. Dec 12, 2008 #4
    Oh that is so easy! I way over thought that. . . Thanks guys
  6. Dec 12, 2008 #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. . .

    Sample mean= 0.8
    St. D= 0.1789
  7. Dec 12, 2008 #6
    "T= (x-bar - mu initial)/(s/ sqr n)

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

    Sample mean= 0.8
    St. D= 0.1789

    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,


    (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.
  8. Dec 14, 2008 #7


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    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 document for more info.

    Hope that helps.

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