Question

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.

Answer 1

Answer:

Step-by-step explanation:

Given that:

A study is conducted and the study by Consumer Reports showed that 64% of supermarket shoppers believe supermarket brands to be as good as national name brands.

a)

Thus; we formulate the null and the alternative hypotheses as follows:

The proportion is ; $\frac{64}{100}$ = 0.64

Null hypothesis:    ${H_0}:p = 0.64$  

The Null hypothesis states that there is no evidence that “the percentage of supermarket shoppers believes that the supermarket ketchup is good as the national brand ketchup differ form 64%”.

Alternative hypothesis: ${H_a}:p \ne 0.64H$

The Alternative hypothesis states that there is evidence that “the percentage of supermarket shoppers believes that the supermarket ketchup is good as the national brand ketchup differ form 64%”.

b)

The proportion of the population = 0.64

The sample size n = 100

The number of shoppers = 52  stating that the supermarket brand was as good as the national brand

The sample proportion $\hat p = \frac{52}{100}$

$\hat p = 0.52$

The z - value is calculated by the formula:

$z = \dfrac{\hat p-p}{\frac{\sqrt{ p(1-p)}}{100 }}$

$z = \dfrac{0.52-0.64}{\frac{\sqrt{ 0.64(1-0.64)}}{100 }}$

$z = \dfrac{-0.12}{0.048 }}$

z = -2.50

Since z = -2.50

the p-value = 2P(Z ≤ -2.50)

p-value = 2 × 0.0062

The p-value = 0.0124

c)

At  significance level,  ∝ = 0.05 ; The p-value = 0.0124

According to  the rejection rule, if p-value is less than 0.05 then we will reject null hypothesis at ∝ = 0.05

Hence, the p-value =0.0124 < ∝ (=0.05)

According to the reject rule; reject null hypothesis.

Conclusion: There is no evidence at all that the percentage of supermarket shoppers believes that the supermarket ketchup is good as the national brand ketchup differ form 64%.

d) we know that the significance level of ∝  = 0.05

The value${z__{0.05}}$  is obtained as:

$P\left( {\left| Z \right| \le z} \right) = 0.05$

Now; to determine Z ; we locate the probability value of 0.05 form the table of standard normal distribution. Then we proceed to the left until the  first column is reached and  the value is 1.90. Also , we move upward until we reach the top and the value is 0.06. Now; the intersection of the row and column results the area  to the left of z

This implies that :$P(Z \leq -1.96)=0.05$

The critical value for left tail is -1.96 and the critical value for right tail is 1.96.

Conclusion:

The critical value is -1.96 and the value of test statistic is - 2.50.  Here, we can see that  the value of test statistic is lesser than the critical value. Hence, we can be concluded that there is evidence that reject the null hypothesis.

Therefore, Yes, the national brand ketchup manufacturer is pleased with the conclusion.