Biostatistics
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 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
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