Question

A study showed that 69% 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 69%.
b. If a sample of 100 shoppers showed 56 stating that the supermarket brand was as good as the national brand, what is the p-value (to 4 decimals)?
c. At α = .05, what is your conclusion?
d. Should the national brand ketchup manufacturer be pleased with this conclusion?

Answer 1

Answer:

We conclude that supermarket ketchup was not as good as the national brand ketchup.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 100

p = 69% = 0.69

Alpha, α = 0.05

Number of stating that the supermarket brand was as good as the national brand , x = 56

a) First, we design the null and the alternate hypothesis

$H_{0}: p = 0.69\\H_A: p \neq 0.69$

This is a two-tailed test.  

b) Formula:

$\hat{p} = \dfrac{x}{n} = \dfrac{56}{100} = 0.56$

$z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

Putting the values, we get,

$z = \displaystyle\frac{0.56-0.69}{\sqrt{\frac{0.69(1-0.69)}{100}}} = -2.810$

Now, we calculate the p-value from the table.

P-value =  0.0049

c) Since the p-value is lower than the significance level, we fail to accept the null and reject it.

Thus, we conclude that supermarket ketchup was not as good as the national brand ketchup.

d) It need to be tested further whether the supermarket brand was worse than the national brand or better than the national brand.