marketing survey using testing population proportion

case study:

A soft drink maker claims that a majority of adults prefer its leading beverage over that of its main competitor’s. To test this claim 500 randomly selected people were given the two beverages in random order to taste. Among them, 270 preferred the soft drink maker’s brand, 211 preferred the competitor’s brand and 19 could not make up their minds. Determine whether there is sufficient evidence, at the 5% level of significance, to support the soft drink maker’s claim against the default that the population is evenly split in its preference. 

solution:

sample_size = 500

sample_p = 271/ sample_size

comp_sample_p = 211/sample_size

alfa = 0.05 # level of significance

#step zero: 

# check if the smaple is large enough

error = sqrt((sample_p * (1 - sample_p))/sample_size)

between(c(sample_p - 3*error, sample_p - 3*error), 0, 1) 

#[1] TRUE TRUE, it means that the sample is large enough

# step one:establishing the test heypothesis:

#H0 population proportion = p0 = 0.500 # claimed proportion for higher popularity

#Ha population proportion > p0 =0.500 # it is a right tailed test

#step two & three: calculating test statistic

z_value = (sample_p - 0.50)/ sqrt((0.5*0.5)/sample_size)

#step 4: finding rejection region/ interval:

#finding the q value

q_value = - qnorm(0.05, mean = 0, sd = 1)

sets::interval(q_value, Inf, "[)")

#[1.64485362695147, Inf)

# step five: decision making 

between(z_value, q_value, Inf)

#[1] TRUE

conclusion: 

the test statistic falls in the rejection region. The decision is to reject H0. In the context of the problem our conclusion is: The data provide sufficient evidence, at the 5% level of significance, to conclude that a majority of adults prefer the company’s beverage to that of their competitor’s.