Topics: Arithmetic mean, Statistical hypothesis testing, Standard deviation Pages: 4 (651 words) Published: March 18, 2013
A. Genetics
Ribosomal 5S RNA can be presented as a sequence of 120 nucleotides. Each nucleotide can be represented by one of four characters: A (Adenine), G (Guanine), C (Cytosine), or U (Uracil). The characters occur with different probabilities for each person. We wish to test if a new sequence is the same as ribosomal 5S RNA. For this purpose, we replicate the new sequence 100 times and find that there are 60 A’s in the20th position. Use a 0.05 level of significance.

1. If the probability of an A in the 20th position is 0.79 in ribosomal 5S RNA, then test the hypothesis that the new sequence is the same as the ribosomal 5S RNA using critical method. 2. Report a p- value corresponding to your result in problem num 1 1A.

* Ho= there is no difference in the new sequence as the ribosomal 5S RNA using critical method Ha= there is difference in the new sequence of ribosomal 5S RNA using critical method

* X= 60, level of significance= 0.05
* Reject null hypothesis
| |
Null Hypothesis =| 0.79|
Level of Significance| 0.05|
Population Standard Deviation| 20|
Sample Size| 100|
Sample Mean| 60|
| |
Intermediate Calculations|
Standard Error of the Mean| 2|
Z Test Statistic| 29.605|
| |
Two-Tail Test|  |
Lower Critical Value| -1.959963985|
Upper Critical Value| 1.959963985|
p-Value| 0|
Reject the null hypothesis|  |
* Conclusion: The new sequence is different from ribosomal 5S RNA using critical method.

* P-value: 0
One method for assessing the effectiveness of a drug is to note its concentration in blood and/or urine sample at certain periods of time after giving the drug. Suppose we wish to compare the concentrations of two types of aspirin (type A and B)in urine specimens taken from the same person, 1hour after he or she has taken the drug. Hence, a specific dosage of either type A or type B aspirin is...
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