Question

In a survey of randomly selected 3,900 family-owned businesses with revenues exceeding $1 million a year, it was found that 1,911 of them had no strategic business plan. a) Use a 90% confidence interval to estimate the proportion of family-owned businesses without strategic business plans. Give an interpretation of this interval. (5 Points) b) Would a 99% confidence interval be wider or narrower than the one you calculated in part (a)

Answer 1

Answer:

a) The 90% confidence interval to estimate the proportion of family-owned businesses without strategic business plans is (0.4768, 0.5032). This means that we are 90% sure that the true proportion of all family-owned businesses without strategic business plans is between these two values.

b) Wider

Step-by-step explanation:

Question a:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

In a survey of randomly selected 3,900 family-owned businesses with revenues exceeding $1 million a year, it was found that 1,911 of them had no strategic business plan.

This means that $n = 3900, \pi = \frac{1911}{3900} = 0.49$

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.49 - 1.645\sqrt{\frac{0.49*0.51}{3900}} = 0.4768$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.49 + 1.645\sqrt{\frac{0.49*0.51}{3900}} = 0.5032$

The 90% confidence interval to estimate the proportion of family-owned businesses without strategic business plans is (0.4768, 0.5032). This means that we are 90% sure that the true proportion of all family-owned businesses without strategic business plans is between these two values.

Question b:

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

The higher the confidence level, the higher the value of z, thus the higher the margin of error and the interval is wider. Thus, a 99% confidence interval is wider than a 90% confidence interval.