Question

A company is conducting a survey of 235 people to measure the level of interest in a new product. Assume that the probability of a randomly selected person’s being "very interested" is 0.88 and that people are selected independently of one another.
a. Find the standard deviation of the percentage who will be found by the survey to be very interested.
b. How much uncertainty is there in the number of people who will be found to be very interested?
c. Find the expected number of people in the sample who will say that they are very interested.
d. Find the expected percentage that the survey will identify as being very interested.

Answer 1

Answer:

a) Standard deviation of the percentage who will be found by the survey to be very interested (SD) =  p(1-p) = 0.88*0.12 = 0.1056

b) By this we mean the margin of error (E) = ± 0.0415

c) Expected number of people  = npq = 235*0.88 = 206.8 => 207 people

d) Expected percentage as being very interest = 0.88*207 = 182.16 = 182 people

Step-by-step explanation:

a) By standard deviation in proportion terms, we mean:

==> σ  = p(1-p). That is the product of the proportion of those being very interested and those otherwise. And since we know the proportion of those very interested to be 0.88. Then q = 1-p => 1 -0.88 = 0.12 not very interested.

Hence, σ  = p(1-p) = 0.88*0.12 = 0.1056.

b) The uncertainty is the margin of error (E) = ± Z*$\sqrt{\frac{p(1-p)}{n} }$, we assumed 95% confidence level and the Z value = 1.96.

Therefore, E = ± 1.96 * $\sqrt{0.1056/235}$ = ± 0.04154838 which we approximated as ± 0.0415.

c) The expected number = np, where n - sample size and p is the proportion of those very interested - since that was the interest in this question.

Therefore,

                E(p) = 235*0.88 = 206.8 = 207 people.

d) Expected percentage as being very interest => The percentage of those very interested multiply by the total number of those expected to be very interested.

And this is: 0.88*207 = 182.16 = 182 people