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ttest helpModerator: BioTeam
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ttest helpWe want to hypothesize that antibacterial solutions work. So we bring 12 petridishes, smear them with isolated bacteria and plant 3 paper discs dipped in antibacterials solutions to see if the bacteria grows or not. 3 days later, we see the bacteria all grown except around the paper discs. We measurure the diameter of the paper disc and now we want to use a ttest to back our hypothesis. How should I do that? what's the p Value? What kind of ttest would we use since we have uneven groups?
Thanks! Dial anti bacterial Dish # Diameter in millimeters 1 :arrow: 15 / 15 / 20 2 :arrow: 20 / 21 / 15 3 :arrow: 21 / 20 / 22 4 :arrow: 28 / 22 / 21 5 :arrow: 25 / 22 / 26 6 :arrow: 20 / 22 / 28 Clorox wipes Dish # Diameter in millimeters 1 :arrow: 20 / 22 2 :arrow: 14 3 :arrow: 10 4 :arrow: 10
At the very least you’d want to test the null hypothesis that the mean diameter of a treated disc is the same as an untreated disc, against the onesided alternative that the diameter of a treated disc is greater than an untreated disc. The pvalue has to be looked up on a table for the number of degrees of fredom for the particuar test. There is no problem with unequal sized groups. There is a version of the test that uses pooled estimates for the standard deviation used to make the tstatistic. I don’t know what your control (nontreated) disc diameter is, unless that is what the four discs of chlorox wipes is supposed to be. Google ttest.
You're not saying this, but I assume that the diameter you're talking about is the diameter of the disc plus any halo surrounding the disc. If the antibiotic works, there should be a halo of clear agar surrounding the disc, where the antibiotic prevented the bacterial lawn from encroaching.
yes we already have controlled environment that is disks dipped in water which has no effect on the bacteria lawn.
The diameter includes the surrounding area around the disc. What should I put in for null hypothesis then?
The null hypothesis states that there is no significant difference between your controls and treatments. Usually this means difference is within 2 SD(95% confidence level).
Living one day at a time;
Enjoying one moment at a time; Accepting hardships as the pathway to peace; ~Niebuhr
The null hypothesis here, is the assertion that the mean diameter of the experimental discs is the same as the mean diameter of the control discs. The only things you “put in” are the means and standard deviations of the two groups when you calculate the tstatistic for the comparison. If you know how to do this for equalsized groups, then you already know how to do it for unequalsized groups. The only difference is the treatment of the standard deviations. You can easily find the formula for both kinds of tests (equal or unequal sized groups) by googling “ttest” or maybe “Student’s ttest”. Once you’ve calculated your t, you have to compare it to a tabulation of tvalues calculated for your degrees of freedom and see whether your observed t is large enough to reject the null hypothesis or not. If it is, then you can reject the null hypothesis of no difference and claim that the antiobiotic had a significant effect.
Mith is saying the same thing. This way is just more formal.
Oh yeah, and you should take blcr's suggestion of using the student curve if your sample size is small.
Living one day at a time;
Enjoying one moment at a time; Accepting hardships as the pathway to peace; ~Niebuhr
The function, =TTEST(array1,array2,tails,type), returns the pvalue for a student's ttest between the data in array1 vs array2. Tails will be either 1 (onetailed or onesided) or 2 (twotailed or twosided) test. Type will be 1 (paired), 2 (equalvariance), or 3 (unequal variance). If you have a calculator with statistical function capability, you can probably do the ttest there, as well. I think a TI83 or better will even calculate the pvalue of the test.
Here are the results I got from a ttest site (http://www.physics.csbsju.edu/stats/ttest.html)
How would I interpret the results? What's my p value and null hypothesis? Student's tTest: Results The results of an unpaired ttest performed at 12:04 on 28NOV2007 t= 2.83 sdev= 4.24 degrees of freedom = 21 The probability of this result, assuming the null hypothesis, is 0.0100 Group A: Number of items= 18 15.0 15.0 15.0 20.0 20.0 20.0 20.0 21.0 21.0 21.0 22.0 22.0 22.0 22.0 25.0 26.0 28.0 28.0 Mean = 21.3 95% confidence interval for Mean: 19.20 thru 23.36 Standard Deviation = 3.86 Hi = 28.0 Low = 15.0 Median = 21.0 Average Absolute Deviation from Median = 2.72 Group B: Number of items= 5 10.0 10.0 14.0 20.0 22.0 Mean = 15.2 95% confidence interval for Mean: 11.25 thru 19.15 Standard Deviation = 5.59 Hi = 22.0 Low = 10.0 Median = 14.0 Average Absolute Deviation from Median = 4.40
The null hypothesis is that the mean disc diameter of Group A = the mean disc diameter of Group B. The observed tstatistic is large enough to reject the null hypothesis at the p=0.01 (or 99%) confidence level. What the plevel is telling you is that there is only a 1% chance of the null hypothesis being correct, given the observed tstatistic. The interpretation is that the antibiotic inhibited the growth of the bacteria as indicated by the significantly larger halo of nongrowth surrounding the treated discs compared to control, nontreated discs. That the 95% confidence intervals about the means don't overlap for the two groups is also evidence that the means are significantly different at the p=0.05 level, but you can reject the null at the stronger, p=0.01 level, so who cares.
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