Module # 6 Assignment

 A.

a. The mean of the population is calculated as the sum of all values divided by the number of values. In this case, it would be (8+14+16+10+11)/5 = 11.8.

b/c.

Sample: (3, 5, 2)

Mean: (3+5+2)/3 = 3.33

Variance: ((3-3.33)^2 + (5-3.33)^2 + (2-3.33)^2) / 3 = 1.56

Standard Deviation: sqrt(1.56) = 1.25

Sample: (3, 5, 1)

Mean: (3+5+1)/3 = 3

Variance: ((3-3)^2 + (5-3)^2 + (1-3)^2) / 3 = 2.67

Standard Deviation: sqrt(2.67) = 1.63

d. Comparing these to the population:

The population mean is (8+14+16+10+11)/5 = 11.8.

The population variance is ((8-11.8)^2 + (14-11.8)^2 + (16-11.8)^2 + (10-11.8)^2 + (11-11.8)^2) / 5 = 9.36.

The population standard deviation is sqrt(9.36) = ~3.06.

B. 

1. The sample proportion p will have approximately a normal distribution if both np and nq are greater than 5. Since p = .95 and q = .05 (since q = 1 - p), we can calculate:

np = .95 * 100 = 95

nq = .05 * 100 = 5 Both np and nq are greater than 5, so yes, the sample proportion p does have approximately a normal distribution.

2. The smallest value of n for which the sampling distribution of p is approximately normal would be the smallest n such that both np and nq are greater than 5. If we take p to be the smaller of the two proportions (p and q), then we need np > 5. Solving for n gives us n > 5/p. So, the smallest n would be the smallest integer larger than 5/p.

Now, let’s calculate the population mean and standard error:

A. Population mean= (8+14+16+10+11)/5 = 11.8

B. Sample of size n= 100

C. Mean of sample distribution: The mean of the sample distribution (also known as the sampling distribution) is equal to the population mean, which is 11.8.

D. Standard Error Qm=Q/square root of n=4.4/square root of 5= The standard error can be calculated as the standard deviation divided by the square root of the sample size. However, we don’t have the standard deviation in this case.

For the table with variables X, x=u, and (x-u)^2, we would need specific values for X and u to calculate these.