Probability, Combination, Normal and Binomial Distribution

1. There are 10 members in a training group. The purpose of the group is to improve the social skills of the members. Before joining the group, each member is given an assessment of their social skills. After attending the group for 2 months, each member is again given another assessment of their social skills. Now, when we compare the assessment before the group and after the group, we find that 9 members have shown improvement and 1 member has shown deterioration.
1. If there were no such training group, the probability of having improvement or deterioration in social skills is 0.5 and is random. What is the probability of observing after 2 months time that all members have improvements in their social skills?

P(all members have improvements) = ₁₀C₁₀(0.5)¹⁰ = 1/1024 = 0.001

2. As (1.1) above, what is the probability of observing after 2 months time that 9 members have social skills improved and 1 member has social skills deteriorated?

P(9 members have improvements) = ₁₀C₉ (0.5)⁹ (0.5)¹= 10/1024 = 0.01

3. If we want to test if the training group is effective, what is the null hypothesis?

H₀: The training group is not effective, i.e. the training group has made no progress, i.e P(training is not effective) = 0.5

4. As (1.3), what is the alternate hypothesis?

H₁: The training group is effective, P > 0.5

5. Using the binomial distribution to test the hypothesis, what would you conclude? Reject or do not reject the null hypothesis? Is the training group effective?

P(having nine or more members showing improvement)

= ₁₀C₉ (0.5)⁹ (0.5)¹ + ₁₀C₁₀(0.5)¹⁰ = 0.011.

Since the probability for having 9 or more members showing improvement is so low (0.011 < 0.05), we can reject the null hypothesis, and conclude that the training group is effective.

2. Look up a Normal-curve table (available at the back of any introduction to statistics text book) and answer the following questions:
1. For a normally distributed random variable “x”, l =10, and r =5, what is the probability of having x>15?

P(x>15) = P(z> (x - l )/r ) = P(z> (15 – 10)/5) = P(z> 1) = 0.5 - 0.3413 = 0.1587.

2. As (2.1) above, what is the probability of having 5<x<15?

P(5<x<15) = P[(5-10)/5<z<(15-10)/5] = P(-1<z<1) = 0.3413 + 0.3413 = 0.6826

3. In a course of Introduction to Social Work, there are two tutors, Mr. Niceguy and Mr. Toughguy. They mark the seminar discussion papers of their tutorial groups separately. Each group has 10 members. The followings are their scores given by the tutors:

1. What is the mean score and standard deviation of score given by Mr. Niceguy? And by Mr. Toughguy?

(See the table above)

2. If we want to standardize the scores given by Mr. Niceguy and Mr. Toughguy to a mean of 55 and standard deviation of 10, what should be the standardized scores for EACH of the students in the two groups?

(See the table above)