Debate and discussion of any biological questions not pertaining to a particular topic.
Mon Nov 19, 2007 3:29 pm
We 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 t-test to back our hypothesis. How should I do that? what's the p Value? What kind of t-test would we use since we have uneven groups?
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
Dish # Diameter in millimeters
1 :arrow: 20 / 22
2 :arrow: 14
3 :arrow: 10
4 :arrow: 10
Mon Nov 19, 2007 5:25 pm
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 one-sided alternative that the diameter of a treated disc is greater than an untreated disc. The p-value 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 t-statistic. I don’t know what your control (non-treated) disc diameter is, unless that is what the four discs of chlorox wipes is supposed to be. Google t-test.
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.
Mon Nov 19, 2007 6:42 pm
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?
Mon Nov 19, 2007 7:28 pm
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).
Mon Nov 19, 2007 7:56 pm
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 t-statistic for the comparison. If you know how to do this for equal-sized groups, then you already know how to do it for unequal-sized 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 “t-test” or maybe “Student’s t-test”. Once you’ve calculated your t, you have to compare it to a tabulation of t-values 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.
Mon Nov 19, 2007 8:07 pm
Oh yeah, and you should take blcr's suggestion of using the student curve if your sample size is small.
Wed Nov 21, 2007 7:11 pm
Do you know how can do the t-test using MS Excel 2007?
Thu Nov 22, 2007 2:04 am
The function, =TTEST(array1,array2,tails,type), returns the p-value for a student's t-test between the data in array1 vs array2. Tails will be either 1 (one-tailed or one-sided) or 2 (two-tailed or two-sided) test. Type will be 1 (paired), 2 (equal-variance), or 3 (unequal variance). If you have a calculator with statistical function capability, you can probably do the t-test there, as well. I think a TI-83 or better will even calculate the p-value of the test.
Wed Nov 28, 2007 6:18 pm
Here are the results I got from a t-test site (http://www.physics.csbsju.edu/stats/t-test.html
How would I interpret the results? What's my p value and null hypothesis?
Student's t-Test: Results
The results of an unpaired t-test performed at 12:04 on 28-NOV-2007
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
Wed Nov 28, 2007 8:23 pm
The null hypothesis is that the mean disc diameter of Group A = the mean disc diameter of Group B. The observed t-statistic is large enough to reject the null hypothesis at the p=0.01 (or 99%) confidence level. What the p-level is telling you is that there is only a 1% chance of the null hypothesis being correct, given the observed t-statistic. The interpretation is that the antibiotic inhibited the growth of the bacteria as indicated by the significantly larger halo of non-growth surrounding the treated discs compared to control, non-treated 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.
Sat Dec 01, 2007 12:23 am
p = 0.01 means 1% probability the results are due chance... it doesn't mean 1% chance null hypothesis being correct...
Sat Dec 01, 2007 1:14 am
Both statements say the same thing.