Answer:
Explanation:
The question is incomplete. The complete question is:
A study by Consumer Reports showed that 64% of supermarket shoppers believe supermarket brands to be as good as national name brands. To investigate whether this result applies to its own product, the manufacturer of a national name-brand ketchup asked a sample of shoppers whether they believed that supermarket ketchup was as good as the national brand ketchup.
a. Formulate the hypotheses that could be used to determine whether the percentage of supermarket shoppers who believe that the supermarket ketchup was as good as the national brand ketchup differed from 64%.
b. If a sample of 100 shoppers showed 52 stating that the supermarket brand was as good as the national brand, what is the p-value?
c. At α = .05, what is your conclusion?
d. Should the national brand ketchup manufacturer be pleased with this conclusion?
Explain.
Solution:
a) We would set up the hypothesis test.
a) For the null hypothesis,
p = 0.64
For the alternative hypothesis,
p ≠ 0.64
b)Considering the population proportion, probability of success, p = 0.64
q = probability of failure = 1 - p
q = 1 - 0.64 = 0.36
Considering the sample,
Sample proportion, P = x/n
Where
x = number of success = 52
n = number of samples = 100
P = 52/100 = 0.52
b We would determine the test statistic which is the z score
z = (P - p)/√pq/n
z = (0.52 - 0.64)/√(0.64 × 0.36)/100 = - 2.5
Recall, population proportion, p = 0.64
The difference between sample proportion and population proportion(p - P) is 0.64 - 0.52 = 0.12
Since the curve is symmetrical and it is a two tailed test, the p for the left tail is 0.64 - 0.12 = 0.52
the p for the right tail is 0.64 + 0.12 = 0.76
These proportions are lower and higher than the null proportion. Thus, they are evidence in favour of the alternative hypothesis. We will look at the area in both tails. Since it is showing in one tail only, we would double the area
From the normal distribution table, the area below the test z score in the left tail 0.00621
We would double this area to include the area in the right tail of z = 2.5 Thus
p = 0.00621 × 2 = 0.012
c) Since alpha, 0.05 > the p value, 0.012, then we would reject the null hypothesis. At 5% significance level, we can conclude that the percentage of supermarket shoppers who believe that the supermarket ketchup was as good as the national brand ketchup differed from 64%.
d) the national brand ketchup manufacturer be pleased with this conclusion because the percentage of supermarket shoppers who believe that the supermarket ketchup was as good as the national brand ketchup might actually be lower than 64%