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Help with T-tests?

  1. Feb 8, 2012 #1
    Hi, I am a little bit confused with a project I am working on. This was a biology lab where we needed to find if there was more mitosis (cell division) occurring on the tip of a root or the shaft of the root.

    Our class means were: 14.7% of cells on the shaft were undergoing mitosis, and 23.1% of cells on the tip were undergoing mitosis.

    We calculated the standard deviation for each mean:
    Tip=8.4
    Shaft=7.8

    And then using a t-test formula, we calculated the t value to be =4.27
    We also calculated the degrees of freedom to be 66.

    Now I just need help analysis the T-test. I don't really know what a value of 4.27 means. Could someone explain it? I know that I am supposed to look at a table of values, but I am still a bit lost. Any help would be appreciated!
     
  2. jcsd
  3. Feb 8, 2012 #2

    chiro

    User Avatar
    Science Advisor

    Hey Loz55 and welcome to the forums.

    Your t-statistic (which is what you are referring to) measures the probability of getting a value as 'extreme' as the result.

    In this context what you are doing is seeing if statistically if not only is one effect bigger than the other, but also if the effects are statistically significantly different.

    Your t-value looks like it is really really high which gives the indication that the two results are highly statistically significantly different, which you can use when providing evidence for the claim that indeed the "tip of the root" has more mitosis than the "shaft of the root".

    The high-value of t corresponds to the huge amount of statistically significant difference.

    In terms of evaluating the actual probability, what you need to realize is that P(T > t) refers to the probability of getting a result as extreme as your t-statistics (4.27). If you pick up statistical tables or use a computer, you'll find that this probability is really really low which means that if the hypothesis that both mitosis processes were not statistically significantly different, then getting this result would be very very unlikely (although it is still possible and its important to realize this).

    What you need to do is to combine your result with a level of 'confidence' and then use that information to provide an argument for your conclusion based on the data.

    If your confidence level is 95% then it means that for a statistic to the right of the mean of the sample distribution (i.e. t-statistic > 0), then the probability being as extreme as it is has to be greater than 2.5% which it is clearly not. From an online calculator the probability is less than 0.001 which means that even using a high interval, it still comes out as giving evidence to reject the hypothesis that there is no statistically significant difference between the two processes.

    So yes in summary, the t-statistic basically encodes a probability that is used to make an inference which in your case refers to the information about how likely it is for your two processes (mitosis on shaft and on tip) are statistically significantly different if they are indeed that way. From this you can then use the information for the means to say which one is 'higher' (which in this case is to do with the mitosis rate).
     
  4. Feb 9, 2012 #3
    Thanks for the quick reply! It was helpful.
     
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