/*! 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 76 Butter Taste Test A student is t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Butter Taste Test A student is tested to determine whether she can tell butter from margarine. She is blindfolded and given small bites of toast that has been spread with either butter or margarine that have been randomly chosen. The experiment is designed so that she will have exactly 15 bites with butter and 15 bites with margarine. She gets 20 right out of 30 trials. Can she tell butter from margarine at a \(0.05\) level of significance? Explain.

Short Answer

Expert verified
No, she can't tell butter from margarine at a 0.05 level of significance. The p-value is greater than the significance level, so we do not reject the null hypothesis that the student is just guessing.

Step by step solution

01

State the hypotheses

The null hypothesis \(H_0\) is that the student is guessing, so the probability of success (i.e., correctly identifying butter or margarine) is 0.5. So, \(H_0: p=0.5\). The alternative hypothesis \(H_a\) is that the probability of success is not 0.5. So, \(H_a: p\neq 0.5\).
02

Calculate test statistic

The test statistic for a binomial distribution is given by \(Z = (X - np_0)/\sqrt{np_0(1-p_0)}\), where X is the number of successes, n is the number of trials, and \(p_0\) is the probability of success under \(H_0\). Here, \(X=20\), \(n=30\), and \(p_0=0.5\). Substituting these values in, we get \(Z = (20 - 30*0.5)/\sqrt{30*0.5*(1-0.5)} = 1.1547\).
03

Calculate p-value

The two-tailed p-value is calculated as \(P(Z>1.1547) + P(Z<-1.1547)\). This can be looked up in the Z-distribution table or calculated using statistical software, and it comes out to be approximately 0.2482.
04

Compare p-value and significance level

Since the p-value (0.2482) is larger than the significance level (0.05), we do not reject the null hypothesis. This means there is not enough statistical evidence to suggest that the student can tell the difference between butter and margarine.

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.

Binomial Distribution
When dealing with experiments or scenarios that have exactly two possible outcomes, such as success or failure, we use the binomial distribution. It's crucial in hypothesis testing when the outcomes are dichotomous.

For instance, in our example of the Butter Taste Test, each bite can only end in a correct identification (success) or an incorrect one (failure). The probability of a success in each trial is the same, and the trials are independent. If a student guesses randomly, this probability should be 0.5 as there are two choices, butter or margarine.

The formula for calculating probabilities in a binomial distribution incorporates the number of trials (\( n \)), the number of successes (\( X \)), and the probability of success (\( p_0 \)), as shown in the solution for the test statistic.
Null Hypothesis
In hypothesis testing, the null hypothesis (\( H_0 \)) is the statement that there is no effect or no difference, and it serves as the assumption we seek evidence against. It's the default position that the observed results are due to chance rather than any significant effect.

In the Butter Taste Test, the null hypothesis posits that the student is merely guessing, translating to her having a 50% chance of identifying whether it's butter or margarine correctly. Essentially, it claims that the student doesn't have an ability that impacts the outcomes beyond random chance. The hypothesis testing framework is designed to challenge this assumption, and evidence leading to its rejection would support the alternate hypothesis (\( H_a \)).
P-value
The p-value is a pivotal concept in statistics used to quantify the evidence against the null hypothesis. It represents the probability of observing results at least as extreme as those in our study if the null hypothesis were true. In simpler terms, it shows how likely the outcome of an experiment can occur by random chance.

In the lesson of the Butter Taste Test, the calculated p-value is obtained based on the test statistic from the binomial distribution and represents the probability of getting 20 or more correct identifications out of 30 under pure guessing. A low p-value suggests that such an outcome is rare if the student was actually guessing, implying that perhaps the student isn't guessing after all.
Statistical Significance
When we test hypotheses, we do so with a threshold in mind that determines whether the results are statistically significant. This threshold is called the significance level, often denoted by alpha (\( \text{alpha} \)) and is subjectively chosen before the analysis. It's a measure of the strength of the evidence needed to reject the null hypothesis.

Commonly used significance levels include 0.05, 0.01, or 0.10. In our Butter Taste Test, the use of a significance level of 0.05 means we require less than a 5% chance that the observed result is due to randomness to reject the null hypothesis. Since the p-value in our test (0.2482) is higher than 0.05, we do not have enough evidence to claim the results are statistically significant, and thus, we do not reject the null hypothesis that the student is guessing.

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

Suppose we are testing people to see whether the rate of use of seat belts has changed from a previous value of \(88 \%\). Suppose that in our random sample of 500 people we see that 450 have the seat belt fastened. Which of the following figures has the correct p-value for testing the hypothesis that the proportion who use seat belts has changed? Explain your choice.

A magazine advertisement claims that wearing a magnetized bracelet will reduce arthritis pain in those who suffer from arthritis. A medical researcher tests this claim with 233 arthritis sufferers randomly assigned either to wear a magnetized bracelet or to wear a placebo bracelet. The researcher records the proportion of each group who report relief from arthritis pain after 6 weeks. After analyzing the data, he fails to reject the null hypothesis. Which of the following are valid interpretations of his findings? There may be more than one correct answer. a. The magnetized bracelets are not effective at reducing arthritis pain. b. There's insufficient evidence that the magnetized bracelets are effective at reducing arthritis pain. c. The magnetized bracelets had exactly the same effect as the placebo in reducing arthritis pain. d. There were no statistically significant differences between the magnetized bracelets and the placebos in reducing arthritis pain.

A Gallup poll conducted in 2017 found that 648 out of 1011 people surveyed supported same-sex marriage. An NBC News/Wall Street Journal poll conducted the same year surveyed 1200 people and found 720 supported same-sex marriage. a. Find both sample proportions and compare them. b. Test the hypothesis that the population proportions are not equal at the 0.05 significance level.

When a person stands trial for murder, the jury is instructed to assume that the defendant is innocent. Is this claim of innocence an example of a null hypothesis, or is it an example of an alternative hypothesis?

Embedded Tutors A college chemistry instructor thinks the use of embedded tutors (tutors who work with students during regular class meeting times) will improve the success rate in introductory chemistry courses. The passing rate for introductory chemistry is \(62 \%\). The instructor will use embedded tutors in all sections of introductory chemistry and record the percentage of students passing the course. State the null and alternative hypotheses in words and in symbols. Use the symbol \(p\) to represent the passing rate for all introductory chemistry courses that use embedded tutors.

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.