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

We need to determine if the belief about the supermarket ketchup is significantly different from 64%. Let the proportion of shoppers who believe that supermarket ketchup is as good as the national brand be denoted by \( p \). The hypotheses are:- Null Hypothesis \( (H_0) \): \( p = 0.64 \) - Alternative Hypothesis \( (H_a) \): \( p eq 0.64 \)

02

Collect Sample Data

From the problem, we have a sample size \( n = 100 \) and 52 out of these 100 shoppers believe the supermarket brand is as good as the national brand. Therefore, the sample proportion \( \hat{p} = \frac{52}{100} = 0.52 \).

03

Standardize the Test Statistic

Use the formula for the test statistic of a proportion:\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1-p_0)}{n}}} \]Substitute the known values:\[ z = \frac{0.52 - 0.64}{\sqrt{\frac{0.64 \cdot (1-0.64)}{100}}} = \frac{-0.12}{\sqrt{0.2304/100}} = \frac{-0.12}{0.048} \approx -2.50 \]

04

Calculate the p-value

A \( z \)-score of \(-2.50\) corresponds to a \( p \)-value. For a two-tailed test, we need to find the probability that \( z \leq -2.50 \) or \( z \geq 2.50 \). This \( p \)-value can be found using a standard normal distribution table or calculator, which gives \( p \approx 0.0124 \).

05

Make a Conclusion at \( \alpha = 0.05 \)

Since the \( p \)-value (0.0124) is less than the significance level \( \alpha = 0.05 \), we reject the null hypothesis \( H_0 \). This indicates that the proportion of shoppers who believe supermarket ketchup is as good as the national brand is significantly different from 64%.

06

Evaluate the Conclusion for the Manufacturer

The national brand ketchup manufacturer should be pleased with the conclusion because the proportion of shoppers who believe their ketchup is equivalent to the supermarket brand is less than the previous study's 64%. This suggests that more people prefer the national brand over the supermarket brand.

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Key Concepts

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

Sample Proportion

In statistical analysis, a sample proportion is a way of summarizing the results from a sample by calculating the ratio of favorable outcomes to the total sample size. For example, in our exercise, the sample consists of 100 shoppers.

• Out of these, 52 shoppers believe that supermarket ketchup is as good as the national brand.
• The sample proportion (\( \hat{p} \)) is calculated as: \( \hat{p} = \frac{52}{100} = 0.52 \).

This proportion shows us the specific part of our sample that holds the belief in question. By understanding sample proportions, we can make inferences about a larger population.
It's crucial to recognize that sample proportions may vary based on the sample size and the actual proportion in the entire population. Larger samples generally provide more accurate estimates of the population proportion, reducing variability and increasing confidence in the results.

Null Hypothesis

The null hypothesis (\( H_0 \)) is a fundamental concept in hypothesis testing. It is a statement that assumes no effect or no difference, serving as the starting point for statistical testing.

• In this exercise, the null hypothesis is that the proportion of shoppers who believe supermarket ketchup is as good as the national brand is equal to 64%, denoted as \( H_0: p = 0.64 \).

This assumption provides a baseline to determine statistical significance.
In hypothesis testing, we process our sample data to see if there is enough evidence to reject the null hypothesis. We use test statistics and \( p \)-values in this evaluation.
Rejection or non-rejection of \( H_0 \) helps us interpret if our data supports the existence of a noticeable deviation from what was previously considered true about the population.

p-value Calculation

The \( p \)-value is a critical component in hypothesis testing. It measures the strength of evidence against the null hypothesis. A low \( p \)-value suggests that the observed data is unlikely under \( H_0 \), leading us to consider alternative hypotheses.

• To find the \( p \)-value, we calculate the test statistic, as seen in the example provided: \( z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1-p_0)}{n}}} \).
• Here, \( z \approx -2.50 \), indicating a significant deviation from the expected population proportion under \( H_0 \).

For a two-tailed test, the \( p \)-value for \( z = -2.50 \) can be found using standard statistical tables or software, yielding a value of approximately 0.0124.
Since this \( p \)-value is less than the significance level (\( \alpha = 0.05 \)), it suggests rejecting the null hypothesis in confidence that the population proportion is indeed different from the initially assumed 64%.