A food processing company markets a soft cheese spread that is sold in a plastic

Question

A food processing company markets a soft cheese spread that is sold in a plastic container with an “easy pour” spout. Although this spout works extremely well and is popular with consumers, it is expensive to produce. Because of the spout's high cost, the company has developed a new, less expensive spout. While the new, cheaper spout may alienate some purchasers, a company study shows that its introduction will increase profits if less than 10 percent of the cheese spread's current purchasers are lost. That is, let p be the true proportion of all current purchasers who would stop buying the cheese spread if the new spout were used, profits will increase as long as p is less than .10.

Suppose 63 of 1,000 randomly selected purchasers say that they would stop buying the cheese spread if the new spout were used. To assess whether p is less than .10, assume for the sake of argument that p equals .10, and use the sample information to weigh the evidence against this assumption and in favor of the conclusion that p is less than .10. Let the random variable xrepresent the number of the 1,000 purchasers who say they would stop buying the cheese spread.

1. Assuming that p equals .10, then x is a binomial random variable with n = ? and p = ?

What does this information indicate?

2. Use the normal approximation to the binomial distribution to compute the probability.

Discuss results in detail. Show all work.

Explanation / Answer

1. n=1000, p =0.1 (Assuming p=0.1)

However, we are told from the survey that 63/1000 people will stop.

Therefore, new p=63/1000 = 0.063 (For normal approximation)

2. By the normal approximation to the binomial distrivution, since n is large,

Mean = n*p = 63

Standard deviation = sqrt(np(1-p)) =7.683

We now need P(X>100)

We convert the distribution to a standard normal distribution

For X>100, Z=(100-63)/7.683 = 4.82

For this Z value, p= 7.178035e-7

This is an incredibly low probability, which means that if the method is implemented, there is an extremely low chance of p being greater than 0.1. Therefore, we can conclude that we can proceed with the new method safely.

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