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 GENETICS| | | | Data| 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. 2A. * P-value: 0 Pharmacology: 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 given at one time and the 1- hour urine concentration is measured.

One week later, after the first aspirin has presumably been cleared from the system, the same dosage of the other aspirin is give to the same person and the 1- hour urine concentration is noted. Because the order of giving the drugs may affect the results, a table of random numbers is used to decide which of the types of aspirin to give first this experiment is performed on 10 people; results are given: Person| Aspirin A 1hr concentration (mg%)| Aspirin B 1hr concentration (mg%)| 1| 15| 13| 2| 26| 20| 3| 13| 10| 4| 28| 21| 5| 17| 17| 6| 20| 22| 7| 7| 5| 8| 36| 30| 9| 12| 7| 10| 18| 11|.

Mean| 19. 20| 15. 60| S. D| 8. 63| 7. 78| 1. What are the appropriate hypothesis 2. What are the appropriate procedure to test these hypotheses 3. Conduct the test measure necessary for the problem mentioned. 4. Suppose an ? level is used for the test. What is the relationshipbetween the decision reached with the test procedure in #3. 5. Give interference. 1. Ho: there is no difference between two type of aspirin given randomly Ha: there is a difference between two type of aspirin given randomly 2. The type of method use in this problem is z-test for two mean. 3. PHARMACOLOGY| | | |.

Data| Hypothesized Difference| 0. 05| Level of Significance| 0. 05| Population 1 Sample| Sample Size| 10| Sample Mean| 19. 2| Population Standard Deviation| 8. 63| Population 2 Sample| Sample Size| 10| Sample Mean| 15. 6| Population Standard Deviation| 7. 78| | | Intermediate Calculations| Difference in Sample Means| 3. 6| Standard Error of the Difference in Means| 3. 674307| Z-Test Statistic| 0. 966169| | | Two-Tail Test| Lower Critical Value| -1. 95996| Upper Critical Value| 1. 959964| p-Value| 0. 33396| Do not reject the null hypothesis| 3. 4. Do not reject the null hypothesis 5.