Question

Marketing companies are interested in knowing the population percent of women who make the majority of household purchasing decisions. Suppose a marketing company did do a survey. They randomly surveyed 200 households and found that in 120 of them, the woman made the majority of the purchasing decisions. We are interested in the population proportion of households where women make the majority of the purchasing decisions. Which distribution should you use for this problem?

Answer 1

Answer:

Normal distribution with mean 0.6 and standard deviation 0.035

Step-by-step explanation:

It's a problem related to Population Proportion

In this, p′ = x / n where x represents the number of successes and n represents the sample size. The variable p′ is the sample proportion and serves as the point estimate for the true population proportion. q′ = 1 – p′

The variable p′ has a binomial distribution that can be approximated by the normal distribution

here, X is the number of “successes” where the woman makes the majority of the purchasing decisions for the household. P′ is the percentage of households sampled where the woman makes the majority of the purchasing decisions for the household.

x = 120

n = 200

p’ = 120/200 = 0.6

The sample proportion will follow the normal distribution. The mean of the distribution being p’ = 0.6

The standard deviation for the distribution is calculated as

Standard Deviation = √[(p’) x (1 – p’)/n]

Standard Deviation = √[(0.6) x (0.4)/200] = √0.0012

Standard Deviation = 0.035