Question

Consider a sampling distribution with

p=0.08 and samples of size n each. Using the appropriate formulas, find the mean and the standard absolute deviation of the sampling distribution of the sample proportion.
A) For a random sample of size n=4000
B) For a random sample of size n=1000
C) For a random sample size ofn=250

Answer 1

The mean of the sampling distribution is 0.08 for all given sample sizes. The standard deviation varies depending on the sample size: it is 0.0043 for n=4000, 0.0086 for n=1000, and 0.0172 for n=250.

Step-by-step explanation:

The question involves finding the mean and standard deviation of the sampling distribution of the sample proportion when the population proportion (p) is 0.08. For a binomial distribution, the mean of the sampling distribution of the sample proportion is equal to the population proportion, so it is also 0.08 for all sample sizes.

The standard deviation of the sampling distribution of the sample proportion (σ_p') is calculated using the formula σ_p' = √[(p(1-p))/n]. Let's calculate it for different sample sizes:

1. For n = 4000: σ_p' = √[(0.08(1-0.08))/4000] = √[(0.08(0.92))/4000] = √[(0.0736)/4000] = 0.0043
2. For n = 1000: σ_p' = √[(0.08(0.92))/1000] = 0.0086
3. For n = 250: σ_p' = √[(0.08(0.92))/250] = 0.0172

In summary, as the sample size increases, the standard deviation decreases, reflecting less variability in the sampling distribution of the sample proportion.