/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 Taste Testing, continued In Exer... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Taste Testing, continued In Exercise 9.37 , we tried to prove that the national brand tasted better than the store brand. \({ }^{10}\) Perhaps, however, the store brand has a better taste than the national brand! If this is true, then the store brand should be judged as better more than \(50 \%\) of the time. a. State the null and alternative hypothesis to be tested. Is this a one- or a two-tailed test? b. Suppose that, of the 35 food categories used for the taste test, the store brand was found to be better than the national brand in six of the taste comparisons. Use this information to test the hypothesis in part a. Use \(\alpha=.01\). c. In the other 21 food comparisons in this experiment, the tasters could find no difference in taste between the store and national brands. What practical conclusions can you draw from this fact and from the two hypothesis tests in Exercises \(9.37(\mathrm{~b})\) and \(9.38(\mathrm{~b}) ?\)

Short Answer

Expert verified
Answer: Yes, there is evidence to suggest that the store brand might have a better taste than the national brand, or at least, they are not consistently worse in taste.

Step by step solution

01

State the null and alternative hypotheses and identify the type of test

The null hypothesis (H0) is that the store brand is judged as better than the national brand 50% of the time, indicating that the two brands have equal taste. The alternative hypothesis (H1) is that the store brand is judged as better than the national brand more than 50% of the time, suggesting that the store brand has better taste. Since we are only interested in whether the store brand scored better than 50% of the time, this is a one-tailed test. Null hypothesis (H0): P = 0.5 Alternative hypothesis (H1): P > 0.5
02

Use the given information to conduct the hypothesis test

We are given that the store brand was found to be better than the national brand in six of the 35 food category taste comparisons, and we want to test this hypothesis at a 0.01 significance level (α = 0.01). Since the sample size is small (n = 35), we will use the binomial test. The probability of success (observing the store brand as better) is 0.5. Thus, the probability of observing six or more successes (store brands judged as better) in 35 trials can be calculated using the binomial distribution: P(X ≥ 6) = Σ [C(35, k) * (0.5)^k * (0.5)^(35-k)] for k = 6 to 35 Where C(35, k) are the binomial coefficients. Calculate the above probability: P(X ≥ 6) ≈ 0.0032 Since P(X ≥ 6) ≈ 0.0032 is less than α = 0.01, we reject the null hypothesis.
03

State the practical conclusions from this hypothesis test and the previous one

The results from this hypothesis test, along with the previous test in Exercise 9.37(b), suggest that there is strong evidence against the equal taste hypothesis. The current test indicates that the store brand is significantly better than the national brand in taste, while the previous test failed to provide evidence that the national brand was better. Additionally, in 21 other food comparisons, tasters could not find any difference in taste between the store and national brands. This information further strengthens the argument that the store brand might have a better taste than the national brand, or at least in some categories, it is not inferior to the national brand as commonly perceived. In conclusion, the results suggest that there is evidence to indicate that the store brand may have a better taste than the national brand, or at least, they are not consistently worse in taste. Consumers should not dismiss store brands as having inferior taste, and they might even prefer them in some food categories.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Null Hypothesis
In statistical hypothesis testing, the null hypothesis (denoted as \( H_0 \)) is a statement that proposes no effect or no difference — it suggests the status quo or a baseline assumption. It is the hypothesis that researchers aim to test against. In the context of the taste testing example provided, the null hypothesis would be: "The store brand and the national brand have an equal chance of being preferred, i.e., the store brand is not better than the national brand." This translates mathematically to \( P = 0.5 \), where \( P \) is the probability that the store brand is judged as better.

When conducting hypothesis testing, the goal is typically to challenge the null hypothesis in hopes of finding new evidence (alternative hypothesis). If evidence strongly refutes \( H_0 \), researchers may discard it in favor of the alternative hypothesis. It is crucial to start with a clear null hypothesis since all computational aspects of hypothesis testing, such as test choice and significance level determination, rely on it.
Alternative Hypothesis
The alternative hypothesis (denoted as \( H_1 \)) is what the researcher wants to prove. It suggests a new outcome or effect — a change from the status quo represented by the null hypothesis. In the context of the provided exercise, the alternative hypothesis is that the store brand is judged as better than the national brand more than 50% of the time, represented mathematically as \( P > 0.5 \).

This hypothesis indicates that the store brand is potentially better in taste compared to the national brand. When testing hypotheses, the null and alternative are contrasting, and only one can be favored based on the evidence. Researchers look for enough statistical evidence that would lead them to believe that the alternative hypothesis could be true, resulting in rejection of the null hypothesis. Importantly, this specific test is one-tailed because we are testing the direction of the effect — whether the store brand is "better than" rather than simply "different from" the national brand.
Binomial Test
A binomial test is a statistical test used to determine if there is a significant difference between the observed proportion of successes in a binary outcome and a theoretical expected proportion. It is especially handy when working with small binary data sets, as seen in our exercise involving 35 taste test comparisons.

In this scenario, we are assessing whether the store brand is judged as better than the national brand in more than half of the cases. Given the test involves "successes" and "failures" (i.e., store brand preferred or not), and considering the small sample size, the binomial test is suitable.

Using the binomial distribution, we can calculate the probability of observing a certain number of successes. Here, we check the probability that the store brand is preferred at least six times out of 35, given the null hypothesis of a 50% chance \((P = 0.5)\). If this probability is very low (below a pre-determined significance level), it suggests that our observation (store brand being better in this case) is unlikely under the null hypothesis.
Significance Level
In hypothesis testing, the significance level \( \alpha \) is the threshold used to determine whether a hypothesis test's results are statistically significant. It is the probability of rejecting the null hypothesis when it is, in fact, true (a type I error). For example, a significance level of \( \alpha = 0.01 \) means there is a 1% risk of concluding that there is an effect when there is none.

In the taste test exercise, a significance level of 0.01 was chosen, indicating very strict criteria for accepting the alternative hypothesis. If the calculated probability (from our binomial test) is less than this significance level, we reject the null hypothesis in favor of the alternative hypothesis. This decision-making rule ensures that we only reject the null hypothesis when there is strong evidence against it, protecting us from incorrect conclusions.

A lower significance level means the test is more selective about claiming that findings are statistically significant. This mitigates the risk of false positives, ensuring the results are more reliable.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Early Detection of Breast Cancer Of those women who are diagnosed to have early-stage breast cancer, one-third eventually die of the disease. Suppose a community public health department instituted a screening program to provide for the early detection of breast cancer and to increase the survival rate \(p\) of those diagnosed to have the disease. A random sample of 200 women was selected from among those who were periodically screened by the program and who were diagnosed to have the disease. Let \(x\) represent the number of those in the sample who survive the disease. a. If you wish to determine whether the community screening program has been effective, state the alternative hypothesis that should be tested. b. State the null hypothesis. c. If 164 women in the sample of 200 survive the disease, can you conclude that the community screening program was effective? Test using \(\alpha=.05\) and explain the practical conclusions from your test.

A random sample of \(n=1000\) observations from a binomial population produced \(x=279 .\) a. If your research hypothesis is that \(p\) is less than .3 , what should you choose for your alternative hypothesis? Your null hypothesis? b. What is the critical value that determines the rejection region for your test with \(\alpha=.05 ?\) c. Do the data provide sufficient evidence to indicate that \(p\) is less than \(.3 ?\) Use a \(5 \%\) significance level.

PCBs Polychlorinated biphenyls (PCBs) have been found to be dangerously high in some game birds found along the marshlands of the southeastern coast of the United States. The Federal Drug Administration (FDA) considers a concentration of PCBs higher than 5 parts per million (ppm) in these game birds to be dangerous for human consumption. A sample of 38 game birds produced an average of 7.2 ppm with a standard deviation of \(6.2 \mathrm{ppm}\). Is there sufficient evidence to indicate that the mean ppm of \(\mathrm{PCBs}\) in the population of game birds exceeds the FDA's recommended limit of \(5 \mathrm{ppm}\) ? Use \(\alpha=.01\).

Suppose you wish to detect a difference between \(\mu_{1}\) and \(\mu_{2}\) (either \(\mu_{1}>\mu_{2}\) or \(\mu_{1}<\mu_{2}\) ) and, instead of running a two-tailed test using \(\alpha=.05,\) you use the following test procedure. You wait until you have collected the sample data and have calculated \(\bar{x}_{1}\) and \(\bar{x}_{2}\). If \(\bar{x}_{1}\) is larger than \(\bar{x}_{2},\) you choose the alternative hypothesis \(H_{\mathrm{a}}: \mu_{1}>\mu_{2}\) and run a one-tailed test placing \(\alpha_{1}=.05\) in the upper tail of the \(z\) distribution. If, on the other hand, \(\bar{x}_{2}\) is larger than \(\bar{x}_{1},\) you reverse the procedure and run a one-tailed test, placing \(\alpha_{2}=.05\) in the lower tail of the \(z\) distribution. If you use this procedure and if \(\mu_{1}\) actually equals \(\mu_{2},\) what is the probability \(\alpha\) that you will conclude that \(\mu_{1}\) is not equal to \(\mu_{2}\) (i.e., what is the probability \(\alpha\) that you will incorrectly reject \(H_{0}\) when \(H_{0}\) is true)? This exercise demonstrates why statistical tests should be formulated prior to observing the data.

Find the appropriate rejection regions for the large-sample test statistic \(z\) in these cases: a. A right-tailed test with \(\alpha=.01\) b. A two-tailed test at the \(5 \%\) significance level

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.