Thread: Need some help with this statistics word problem

Sorry forgot to add sample n is 1419 and population is 30,000

Originally Posted by TheBigSmoke

I don't even need the answer, I just want to a understand where to start with this question and what formula to use, I've been thinking about it for hours and there aren't even any examples in my textbook of a question like this, it isn't estimation or hypothesis from what I can tell... here is the question sorry it a bit long and thanks in advance for any help at all..


One criticism, and myth, of entrepreneurship is that the majority of new businesses fail. This
simply isn’t true. The often-used statistic that 9 out of 10 businesses fail in their first few years
is an exaggeration. According to the US Small Business Administration, after 4 years it is
estimated that 50% of new businesses are successful and still operating, 17% are closed (but
were considered to be successful by their owners) and 33% were considered to be out and out
failures. The 33% of businesses, classified as failures indicate that motivation to start and run a
business simply isn’t enough – in other words, if you build it, doesn’t mean they’ll come.
Consequently, it’s of considerable importance to understand the factors (beyond motivation) that
may contribute to improving the odds that a new business will become successful.

One success factor is believed to be the entrepreneur’s utilization of Ontario’s RIC (Regional
Innovation Centres) program – provincially funded business advisory centres. There are now
14 RIC’s in Ontario. Another factor is believed to be the experience level of the entrepreneur. It
is believed the # of hours of experience an entrepreneur needs to become truly competent and
proficient in any given field takes, on average, 10,000 hours. This represents about 5 years of
post-school ‘apprenticeship’. Ministry officials believe that more than 25% of their entrepreneur ‘clients’ have more than 10,000 hours of experience.

P-hat = 324/1419 = 0.2283, a = 0.05. sample mean = 6491, sample SD = 3747.5606

THE QUESTION

a) Exactly how many clients do they believe will have this experience level?

b) How many clients (# and proportion) actually have this experience?

Neat problem.

a) This is about finding a 95% confidence interval for $\hat{p}$ and using the high end of it to make their claim.

you obtain the sample standard deviation as $\sigma_{\hat{p}}=\left(\dfrac{\hat{p}(1-\hat{p})}{N_{sample}}\right)^\frac{1}{2}$

then the distribution for $\hat{p}$ is Normal($\hat{p},\sigma_{\hat{p}})$ and from that you can construct a confidence interval for $\hat{p}$

If you do this you'll find the 95% confidence interval for $\hat{p} = (0.20649, 0.25017)$ and thus their claim of 25% just makes the cut of being true.

I guess they believe that they have $0.25017 \cdot N_{pop} = 0.25017 \cdot 30000 = 7505$


b) This is about using the sample statistics to find the how many clients actually have more than 10K hrs experience. The distribution for experience is Normal(6491, 3747.5606) and you can easily find the probability someone has >10k hrs. This turns out to be about 17.5%

So the proportion of the population that have >10k hrs of experience is 17.5% and thus the number of the population that actually have this experience is

$.175 \cdot 30000 = 5236$

This problem is a nice example of "statistics in action".