Question

1. Given the following data, compute the mean marks and standard deviation

Class       0-25   25-50  50-75   75-100

Frequency 30 50 80 40

2. Explain the importance and application of Linear Regression in drug stability studies.

3. Perform the one-way ANOVA for the data of testing of a new drug on four patients, data is given below:

Clinic 1 2 3 4 5 6 7

Patient 1 5 7 12 9 11 6 10

Patient 2 3 7 10 8 9 9 10

Patient 3 6 7 5 7 13 9 7

Patient 4 8 11 6 5 7 12 9

Answer 1

1.

Mean $= \frac{\sum fx}{\sum f} = \frac{10750}{200} = 53.75$

Standard deviation $= \sqrt{ \frac{\sum fx^2}{\sum f} - (\frac{\sum fx}{\sum f})^2 }$

$= \sqrt{ \frac{693750}{200} - (\frac{10750}{200})^2 } \\ = \sqrt{ \frac{693750}{200} - (\frac{10750}{200})^2 } \\ = \sqrt{ 3468.75 -2889.0625 } \\ = 24.076$

2. Evaluation of stability data of drug substances and products is required to perform a statistical extrapolation of a shelf-life of a drug product or a retest period for a drug substance is based heavily on whether data exhibit a change-over-time and/or variability. We can apply simple statistical parameters for determining whether sets of stability data from either accelerated or long-term storage programs exhibit a change-over-time and/or variability. These parameters are all derived from a simple linear regression analysis first performed on the stability data.

3.