91Ó°ÊÓ

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. \(\quad\) At \(\alpha=.05,\) what is your conclusion? d. Should the national brand ketchup manufacturer be pleased with this conclusion? Explain.

Step by step solution

01

Formulate Hypotheses

To determine whether the percentage of supermarket ketchup believers differs from 64%, we set up the hypotheses as follows:\[H_0: p = 0.64 \H_a: p eq 0.64\]where \(p\) is the true proportion of supermarket shoppers that believe supermarket ketchup is as good as the national brand.

02

Calculate the Sample Proportion

Given that 52 out of 100 shoppers believe the supermarket brand is as good as the national brand, we calculate the sample proportion \( \hat{p} \):\[\hat{p} = \frac{52}{100} = 0.52\]

03

Compute the Test Statistic

We use the test statistic for a proportion, defined as:\[z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\]Substituting the known values:\[z = \frac{0.52 - 0.64}{\sqrt{\frac{0.64 \times 0.36}{100}}} = \frac{-0.12}{0.048} = -2.5\]

04

Find the p-value

Using the standard normal distribution, we find the p-value for \(z = -2.5\). The p-value is:\[p = 2 \times P(Z < -2.5) \approx 2 \times 0.0062 = 0.0124\]

05

Make a Decision

Compare the p-value to the significance level \(\alpha = 0.05\). Since \(p = 0.0124 < 0.05\), we reject the null hypothesis. This suggests that the proportion of all supermarket shoppers who believe supermarket ketchup is as good as the national brand differs from 64%.

06

Conclusion for the Manufacturer

The rejection of the null hypothesis indicates that fewer than 64% of shoppers believe supermarket ketchup is as good as the national brand. Therefore, the manufacturer should be pleased with this conclusion, as their product is perceived relatively better.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Proportion Testing

Proportion testing plays a vital role in statistics when you aim to understand if a specific proportion within a population differs significantly from a stated norm or benchmark. In the given exercise, the focus is on comparing the proportion of people who think supermarket ketchup is as good as the national brand, with an established proportion of 64%.
The process involves:

• Defining the sample – in this case, 100 shoppers.
• Calculating the sample proportion, which is the fraction of the sample that meets the criteria – here, 52 out of 100.
• Determining if this observed sample proportion deviates significantly from the claimed proportion of 64%.

To achieve this, hypothesis testing is employed where the null hypothesis (H0) usually states no difference, while the alternative hypothesis (Ha) suggests a deviation exists. This foundational concept allows businesses and researchers to make educated decisions based on observed data.

p-value

The p-value is a cornerstone concept in hypothesis testing. It helps us determine the strength of the evidence against the null hypothesis. In simple terms, it quantifies the probability of obtaining an observed sample statistic, like our sample proportion of 0.52, if the null hypothesis were true.
For the ketchup manufacturer case, the p-value is 0.0124, which indicates a very small probability of observing such a deviation (or a greater one) if 64% of shoppers truly believed supermarket ketchup was as good as the national brand.
A lower p-value indicates stronger evidence against the null hypothesis. In the example, a p-value below the chosen significance level of 0.05 suggests that the difference in proportions is statistically significant. This informs researchers whether to reject the null hypothesis confidently or not.

Significance Level

The significance level, often denoted as \(\alpha\), is the threshold at which you judge the p-value. It represents the probability of wrongly rejecting the null hypothesis when it is actually true, known as a Type I error.
Common significance levels are 0.05, 0.01, and 0.10. In our exercise, a significance level of 0.05 is used. This means that there is a 5% risk of concluding that a difference exists when there is none.
When the p-value (0.0124) is less than the significance level (0.05), as in the case of the ketchup study, it suggests that the null hypothesis should be rejected. This criterion allows researchers to control how much statistical evidence is needed to assert that a true effect exists.

Null Hypothesis

The null hypothesis (H0) is a fundamental component of hypothesis testing. It embodies the idea that there is no effect or difference, acting as a baseline or starting assumption.
In this ketchup analysis, the null hypothesis states that the proportion of shoppers who believe that supermarket ketchup is as good as the national brand is indeed 64% (H0: p = 0.64). The analysis sets out to test this assumption.
Rejecting the null hypothesis, as was the result in this exercise, implies finding sufficient statistical evidence that the proportion is different from 64%. The outcome suggests that perceptions of the ketchup might not be as positive, which has strategic implications for the national brand manufacturer.
Understanding the null hypothesis is essential because it defines the direction and interpretation of your statistical tests.