Sampling distribution

Question

It is reported that 63% of all entrepreneurs who opened new enterprises in a certain country in the previous year were university graduates. Suppose a random sample of 300 entrepreneurs who opened new enterprises in the previous year is selected in order to learn about which industries are most appealing to entrepreneurs.

REQUIRED:

(i) What is the probability that the sample proportion of enterprises opened by university graduates will be greater than 0.6?

(ii)What is the probability that the sample proportion of enterprises opened by university graduates will be less than 0.7?

(iii) Develop and interpret a 94% confidence interval estimate for the proportion of enterprises opened by university graduates.

Answer 1

The standard error for sample proportions is

SE(p-hat) = √[p(1-p)/n] = √0.63*0.37/300 =

0.0278747197295

The mean for sample proportions, µ, is the population proportion, p = 0.63

We can standardize(normalize) the condition p-hat > 0.6, by subtracting the mean and dividing by the SE.

p-hat - µ > 0.6 - 0.63

(p-hat - µ)/σ > (0.6 - 0.63)/0.02787 = −1.076244005

The left side is the normal statistic z, so the condition p-hat > 0.6 has been “normalized” now to z > -1.08 (rounding for later use on z-table), and now we can find the probability of this happening.

I’ll stop here for now, but let me know if you would like help with the rest.