Question

There are 2567 students enrolled at a small college, with 2053 of them enrolled in a sociology course. In the sampling distribution of sample proportions of size 230, above what proportion will 52% of all sample proportions be

a) 0.52
b) 0.89
c) 0.76
d) 0.42

Answer 1

To find the proportions above which 52% of all sample proportions will be, we need to calculate the margin of error and subtract it from the given proportions. For the given proportions of 0.52, 0.89, 0.76, and 0.42, the proportions above which 52% of all sample proportions will be are 0.4407, 0.8562, 0.7179, and 0.3633 respectively.

Step-by-step explanation:

In order to answer the question, we need to find the critical value for a given confidence level.

For a 95% confidence level, the critical value is 1.96.

Now, we can calculate the margin of error (ME) using the formula:

ME = critical value * standard deviation

Given that the standard deviation is the square root of (proportion * (1 - proportion)), we can calculate the margin of error (ME) for each proportion:

a) For 0.52 proportion:

ME = 1.96 * sqrt(0.52 * (1 - 0.52))

ME = 0.0793

b) For 0.89 proportion:

ME = 1.96 * sqrt(0.89 * (1 - 0.89))

ME = 0.0338

c) For 0.76 proportion:

ME = 1.96 * sqrt(0.76 * (1 - 0.76))

ME = 0.0421

d) For 0.42 proportion:

ME = 1.96 * sqrt(0.42 * (1 - 0.42))

ME = 0.0567

Therefore, for a sample proportion to be above a given proportion, the margin of error must be subtracted from the given proportion. Here are the results:

a) For a proportion of 0.52: 0.52 - 0.0793 = 0.4407

b) For a proportion of 0.89: 0.89 - 0.0338 = 0.8562

c) For a proportion of 0.76: 0.76 - 0.0421 = 0.7179

d) For a proportion of 0.42: 0.42 - 0.0567 = 0.3633