Question

Suppose the population proportion of American citizens who are in favor of gun control is .61. If a sampling distribution of size n= 50 was created from this population, what would be the mean of this sampling distribution?

Answer 1

The Mean of the sampling distribution is μ = p = 0.61

Explanation:

Given size of the sampling distribution (n) = 50

Suppose the population proportion of American citizens who are in favor of gun control is .61

That is p = 0.61

Sampling distribution of proportions:-

Let p be the probability of occurrence of an event (called its success) and q =1-p is the probability of non- occurrence (called its failure).Draw all possible samples of size n from an infinite population.

Compute the proportion P of success for each of these samples. Then
`u_(p)`

and variance sampling distributions are given by

`u_(p) =p` and

variance
`(pq)/(n)`

Standard deviation (S.D) =
`\sqrt{(pq)/(n)}`

Mean of the sampling distribution:-

The Mean of the sampling distribution is μ = p

Given data the proportion of American citizens who are in favor of gun control is 0.61

p = 0.61

The Mean of the sampling distribution is μ = p = 0.61