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In problem \(8.16\), a college chemistry instructor thinks the use of embedded tutors will improve the succes: rate in introductory chemistry courses. The instructor carried out a hypothesis test and found that the observed value of the test statistic was \(2.33 .\) The \(\mathrm{p}\) -value associated with this test statistic is \(0.0099 .\) Explain the meaning of the p-value in this context. Based on this result, should the instructor believe the success rate has improved?

Short Answer

Expert verified
The p-value in this context, 0.0099, indicates a very low probability (less than 1%) of observing the results if the null hypothesis is true (i.e., there was no difference or improvement in the success rate). Since this p-value is less than the conventional threshold of 0.05, it implies strong evidence against the null hypothesis. Therefore, based on this result, the instructor should believe the success rate has improved with the introduction of embedded tutors.

Step by step solution

01

Understanding Test Statistic and P-value

In hypothesis testing in statistics, the test statistic is used to make a decision about the null hypothesis which states that there is no effect or difference. The p-value is the probability of obtaining results as extreme as the observed results of a statistical hypothesis test, assuming the null hypothesis is true. The smaller the p-value, the stronger the evidence against the null hypothesis. Typically, a p-value less than 0.05 is considered statistically significant, providing strong evidence against the null hypothesis.
02

Interpreting the P-value

In this context, the p-value of 0.0099 is less than 0.05, which suggests the likelihood of accidentally rejecting the null hypothesis when it is true is very low. That means, the difference observed in the success rate is likely not due to random chance and there's strong evidence against the hypothesis that there's no difference.
03

Final Decision

Based on the interpretation of the p-value, the instructor should conclude there's strong evidence in support of the alternative hypothesis - that implies the use of embedded tutors has indeed improved the success rate in introductory chemistry courses.

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Key Concepts

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

Hypothesis Test
A hypothesis test in statistics is a method used to evaluate two mutually exclusive statements about a population to determine which statement is best supported by the sample data. It begins with an assumption called the null hypothesis (ull hypothesis), typically representing the status quo or a position of no change or no effect.

For example, the college chemistry instructor used a hypothesis test to determine whether embedded tutors affected the success rate of students. The hypothesis test framework allowed the instructor to systematically make an inference about the population of chemistry students based on the observed sample data.
Test Statistic
The test statistic is a calculated value that results from running a statistical test. This value allows us to decide whether to accept or reject the ull hypothesis. The test statistic is calculated from the sample data and is used to assess the likelihood of a given hypothesis. It is a part of a hypothesis test and is generally based on a standardized scale so that the result can be interpreted.

In our problem, the test statistic of 2.33 suggests that if the null hypothesis is true, obtaining a result as extreme or more extreme than 2.33 is unlikely. Thus, test statistics guide us toward a more concrete decision-making process in hypothesis tests.
Null Hypothesis
The null hypothesis (ull hypothesis) is the default assumption that there is no relationship, effect, or difference between certain conditions or variables. In hypothesis testing, it represents the hypothesis that there is no significant difference or the hypothesis we seek evidence against.

In the context of the chemistry class, the null hypothesis would be that the introduction of embedded tutors has no positive effect on the success rate. This is the premise the instructor is testing against.
Statistical Significance
Statistical significance is a term used to determine if the evidence in the sample data is strong enough to draw conclusions about the population. It suggests how likely it is that the obtained result would occur if the null hypothesis were true and generally indicates a low probability of the observed data occurring by random chance.

A p-value lower than the threshold (commonly 0.05) is an indication of statistical significance. In the given problem, since the p-value is 0.0099, the findings are considered to be statistically significant, and it can be argued with confidence that the results were not due to random chance but likely due to the use of embedded tutors.
Alternative Hypothesis
The alternative hypothesis (ull hypothesis) is the opposing statement to the null hypothesis in a hypothesis test. It represents the outcome that the study is attempting to prove—in our case, that the embedded tutors do have a positive impact on students' success rates. The alternative hypothesis is accepted when evidence against the null hypothesis is strong.

With a p-value of 0.0099, which is below the commonly accepted significance level, the instructor has enough evidence to reject the null hypothesis in favor of the alternative hypothesis. This would indicate that the instructor’s belief that the success rate has improved with the inclusion of embedded tutors is statistically supported.

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Most popular questions from this chapter

If we do not reject the null hypothesis, is it valid to say that we accept the null hypothesis? Why or why not?

In the Pew Research social media survey, television viewers were asked if it would be very hard to give up watching television. In \(2002,38 \%\) responded yes. In \(2018,31 \%\) said it would be very hard to give up watching television. a. Assume that both polls used samples of 200 people. Do a test to see whether the proportion of people who reported it would be very hard to give up watching television was significantly different in 2002 and 2018 using a \(0.05\) significance level. b. Repeat the problem, now assuming the sample sizes were both 2000 . (The actual sample size in 2018 was \(2002 .\).) c. Comment on the effect of different sample sizes on the p-value and on the conclusion.

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A true/false test has 50 questions. Suppose a passing grade is 35 or more correct answers. Test the claim that a student knows more than half of the answers and is not just guessing. Assume the student gets 35 answers correct out of \(50 .\) Use a significance level of \(0.05 .\) Steps 1 and 2 of a hypothesis test procedure are given. Show steps 3 and 4, and be sure to write a clear conclusion. Step $$\text { 1: } \begin{aligned}&\mathrm{H}_{0}: p=0.50 \\\&\mathrm{H}_{\mathrm{a}}: p>0.50\end{aligned}$$ Step 2: Choose the one-proportion \(z\) -test. Sample size is large enough, because \(n p_{0}\) is \(50(0.5)=25\) and \(n\left(1-p_{0}\right)=50(0.50)=25\), and both are more than \(10 .\) Assume the sample is random and \(\alpha=0.05\).

According to the Bureau of Labor Statistics, \(10.1 \%\) of Americans are self- employed. A researcher wants to determine if the self-employment rate in a certain area is different. She takes a random sample of 500 working residents from the area and finds that 62 are self-employed. a. Test the hypothesis that the proportion of self-employed workers in this area is different from \(10.1 \%\). Use a \(0.05\) significance level. b. After conducting the hypothesis test, a further question one might ask, "What proportion of workers in this area are self-employed?" Use the sample data to find a \(95 \%\) confidence interval for the proportion of workers who are self-employed in the area from which the sample was drawn. How does this confidence interval support the hypothesis test conclusion?

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