Question

Karen is interested in starting her own business. Before she begins, she wishes to conduct research on the success of small businesses in the first 2 years. Assume that the percentage of small business owners that turn a profit in the first 2 years is 18%. For what sample size, n, will the sampling distribution of sample proportions have a standard deviation of 0.09?

Answer 1

To find the sample size needed for the sampling distribution of sample proportions to have a standard deviation of 0.09, we can use the formula: n = (z * p * (1-p)) / E^2. Plugging in the values, we get: n ≈ 475.

Step-by-step explanation:

To find the sample size needed for the sampling distribution of sample proportions to have a standard deviation of 0.09, we can use the formula:

n = (z * p * (1-p)) / E^2

Where n is the sample size, z is the z-score corresponding to the desired level of confidence (e.g., for a 95% confidence level, z = 1.96), p is the estimated proportion (0.18 in this case), and E is the desired margin of error (0.09 in this case).

Plugging in the values, we get:

n = (1.96 * 0.18 * (1-0.18)) / 0.09^2

n = 3.834 / 0.0081

n ≈ 474.81

So, the sample size needed for the sampling distribution of sample proportions to have a standard deviation of 0.09 is approximately 475.

Answer 2

The sample size n required for the sampling distribution of sample proportions to have a standard deviation of 0.09, with a population proportion of 0.18, is calculated to be 1825.

Step-by-step explanation:

Karen is interested in starting her own business and wants to research the success of small businesses in their first 2 years. Given that the percentage of small business owners that turn a profit in the first 2 years is 18%, she wants to determine the necessary sample size n such that the sampling distribution of sample proportions has a standard deviation of 0.09.

To find the sample size n, we'll use the standard deviation formula for the sampling distribution of a proportion, which is:

`\[ \sqrt{(p(1-p))/(n)} = 0.09 \]`

Where p is the population proportion (0.18 in this case), and n is the sample size. Rearranging and solving for n gives us:

`\[ n = (p(1-p))/((0.09)^2) \]`

Substituting the given values:

`\[ n = (0.18(1-0.18))/((0.09)^2) \]`

`\[ n = (0.18(0.82))/(0.0081) \]`

`\[ n = (0.1476)/(0.0081) \]`

`\[ n = 1824.69 \]`

Since we cannot survey a fraction of a person, we would round up to the nearest whole number, which gives us n = 1825.