Question

Suppose 5% of students are veterans and 149 students are involved in sports. How unusual would it be to have no more than 20 veterans involved in sports? (20 veterans is about 13.4228%) When working with samples of size 149, what is the mean of the sampling distribution for the proportion of veterans? 0.01785 Х When working with samples of size 149, what is the standard error of the sampling distribution for the proportion of veterans? Compute PCÔ < 0.134228). Pp < 0.134228) = NOTE: Give results accurate to 5 decimal places

Answer 1

We would be looking at the first question all parts here as:

Q1) a) The mean of the sampling distribution for the proportion is computed here as:

$\mu_p = 0.05$

b) The standard error of sampling distribution for the proportion here is computed as:

$SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.05*0.95}{149}} = 0.01785$

Therefore 0.01785 is the required standard error here.

c) The probability here is computed as:

P( p <= 0.134228)

Converting it to a standard normal variable, we have here:

$P(Z < \frac{0.134228 - 0.05}{0.01785})$

Getting it from the standard normal tables, we have here:

Therefore 1.00000 is the required probability here.