Question

After an extensive advertising campaign, the manager of a company expects the proportion of potential customers that recognize a new product to be at least 60%. She samples 120 potential consumers from the population of 1600 potential customers, and finds that 54 recognize this product. What is the probability of getting the sample proportion that she obtained (or one that is lower) if the true population proportion is .60? What type of problem is this? Poisson Random variable Sample mean Sample proportion Normal

Answer 1

Answer:

Step-by-step explanation:

Hello!

The study variable is:

X: number of customers that recognize a new product out of 120.

There are two possible recordable outcomes for this variable, the customer can either "recognize the new product" or " don't recognize the new product". The number of trials is fixed, assuming that each customer is independent of the others and the probability of success is the same for all customers, p= 0.6, then we can say this variable has a binomial distribution.

The sample proportion obtained is:

p'= 54/120= 0.45

Considering that the sample size is large enough (n≥30) you can apply the Central Limit Theorem and approximate the distribution of the sample proportion to normal: p' ≈ N(p;$\sqrt{\frac{p(1-p)}{n} }$)

The other conditions for this approximation are also met: (n*p)≥5 and (n*q)≥5

The probability of getting the calculated sample proportion, or lower is:

P(X≤0.45)= P(Z≤$\frac{0.45-0.6}{\sqrt{\frac{0.6*0.4}{120} } }$)= P(Z≤-3.35)= 0.000

This type of problem is for the sample proportion.

I hope this helps!