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The mean of each vector will be equal to:

The standard deviation of each vector will be equal to:

The standard error of each vector will be equal to:

The formula for the t test is as follows:

The t-test is used to test the difference between the means of two test sets, as in before and after studies or matched-pairs studies.There is a confidence interval for the mean and a critical value for t for the chosen level of significance associated with the t-test. For instance, a level of significance equal to0.05 means that 95% of the cases will be within the confidence range if there is no significant difference between the means of the two test sets, or experiments,being compared. The confidence limits set upper and lower bounds on an estimate of the mean for the chosen level of significance (0.05). The confidence interval is therange within the bounds of the confidence limits. The confidence interval can be computed, if you know the shape of your distribution. For normally distributed data, theconfidence limits at the 0.05 significance level for an estimated mean are the sample mean plus or minus 1.96 times thestandard error.

confidence interval (normal distribution): mean +/- 1.96 * SE

For example, if the sample mean is 10 and the standard error is 1.2, then 95% of the cases will be within the range of 10 plus or minus 1.96 times 1.2,or 10 plus or minus 2.4, which is the range from 7.6 to 12.4. Thus, if the experimental mean is outside the limits of this range computed for the reference mean, then the differencebetween the means of the two test sets is considered to be significant within a probability of 95%. The critical value for t at a given significance level fora specific type of distribution can be looked up in a table; most statistics books contain them. In the case of microarray data, if the absolute value of tis greater than the critical value, this indicates a significant difference in the gene expression between the reference and experimental test sets.Because the t-test is a parametric test that assumes a normal distribution, the statistical tests that are commonly used to analyze microarray data are more complexvariations that are used for distributions other than normal distributions.

Tutorial data set
red:green ratio red:green ratio red:green ratio red:green ratio red:green ratio red:green ratio red:green ratio
Gene measurement 1 measurement 2 measurement 3 measurement 4 measurement 5 measurement 6 measurement 7
A(ref) 0.97 1.54 1.32 0.89 1.06 1.21
A(exp) 1.37 1.25 1.15 0.99 1.30 1.53 1.07
B(ref) 1.67 1.78 2.01 1.89 1.75 1.81 1.69
B(exp) 6.21 6.03 5.94 6.14 6.11

Assumptions for example problem:

  • Use the example data set to perform a t-test analysis.
  • Consider each row of the table a separate test set under either experimental or reference conditions.
  • Assume a normal distribution.
  • Choose 0.05 as the level of significance.
  • Compute answers to the second decimal place.
  • Assume the critical value for t is equal to an absolute value of t greater than 2.37.

What are the means for each row of data?

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What are the standard deviations for each row of data?

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What is the standard error for each row of data?

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What is the value for t for the comparison between the reference and experimental test sets for Gene A?

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What is the 95% confidence interval computed for the Gene A reference set?

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Is there a significant difference between the mean values of the experimental versus the reference set for Gene A? (Explain the answer both in terms of the t value and the confidence interval.)

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What is the value for t for the comparison between the reference and experimental test sets for Gene B?

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What is the 95% confidence interval computed for the Gene B reference set?

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Is there a significant difference between the mean values of the experimental versus the reference set for Gene B? (Explainthe answer both in terms of the t value and the confidence interval.)

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If there was a significant difference between the gene expressionunder experimental conditions versus the gene expression under reference conditions for either Gene A or Gene B, then estimate the significant increaseor decrease observed.

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There are many software packages available that have been designed expressly for microarray data analysis. In addition to testing gene expression under aset of experimental conditions versus reference conditions, it is possible to identified "clustered" genes that seem to have similar responses under similarconditions. Also, genes can be identified that show related responses under similar conditions, such as one gene's expression always increases whenanother's decreases. When two or more genes show this kind of clustered behavior, it can be an indication that they are part of the same pathway,or that they are regulating each other. Using this type of microarray data analysis, the scientist can combine the cluster analysis results with what is known through laboratory experiments and often come up with new hypotheses about biochemical pathways and regulation.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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cm
tijani
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Can you compute that for me. Ty
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what is inorganic
emma
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can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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answer
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progressive wave
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Bios 533 bioinformatics. OpenStax CNX. Sep 24, 2008 Download for free at http://cnx.org/content/col10152/1.16
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