/*! 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 11 Thirty students volunteer to tes... [FREE SOLUTION] | 91Ó°ÊÓ

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Thirty students volunteer to test which of two strategies for taking multiple- choice exams leads to higher average results. Each student flips a coin, and if heads, uses Strategy A on the first exam and Strategy \(\mathrm{B}\) on the second, while if tails, uses Strategy B on the first exam and Strategy A on the second. The average of all 30 Strategy A results is then compared to the average of all 30 Strategy B results. What is the conclusion at the \(5 \%\) significance level if a two- sample hypothesis test, \(H_{0}: \mu_{1}=\mu_{2}, H_{\mathrm{a}}: \mu_{1} \neq \mu_{2},\) results in a \(P\) -value of \(0.18 ?\) (A) The observed difference in average scores is significant. (B) The observed difference in average scores is not significant. (C) A conclusion is not possible without knowing the average scores resulting from using each strategy. (D) A conclusion is not possible without knowing the average scores and the standard deviations resulting from using each strategy. (E) A two-sample hypothesis test should not be used here.

Short Answer

Expert verified
B: The observed difference in average scores is not significant.

Step by step solution

01

- Understand the Null and Alternative Hypotheses

The null hypothesis ( H_0 ) states that the two means are equal: H_0: μ_1 = μ_2 . The alternative hypothesis ( H_a ) states that the two means are not equal: H_a: μ_1 ≠ μ_2 .
02

- Significance Level

The significance level is set at 5%, or 0.05. This means that there is a 5% chance of rejecting the null hypothesis when it is actually true.
03

- Given P-value

The P-value given is 0.18. This represents the probability of obtaining test results at least as extreme as the ones observed during the study, assuming the null hypothesis is true.
04

- Compare P-value to Significance Level

Compare the P-value (0.18) to the significance level (0.05). If the P-value is greater than the significance level, we fail to reject the null hypothesis.
05

- Decision

Since the P-value (0.18) is greater than the significance level (0.05), we fail to reject the null hypothesis. This means that there is not enough evidence to suggest a significant difference between the average scores of the two strategies.
06

- Conclusion

Given that we fail to reject the null hypothesis, we conclude that the observed difference in average scores is not significant at the 5% significance level. This matches option (B).

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Key Concepts

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

Null Hypothesis
The null hypothesis, often denoted as \( H_0 \), is a fundamental part of hypothesis testing. It represents the default or status quo assumption that there is no effect or no difference between groups being compared. In the context of the exercise, the null hypothesis \( H_0: \mu_1 = \mu_2 \) states that there is no difference in the average results between Strategy A and Strategy B. This means we initially assume both strategies produce the same average scores.
By testing the null hypothesis, we determine whether observed data provides enough evidence to conclude that there is a difference. If evidence is not strong enough, we fail to reject the null hypothesis, holding to the presumption that no difference exists between the two strategies.
Alternative Hypothesis
In hypothesis testing, the alternative hypothesis, denoted as \( H_a \), contrasts the null hypothesis. It suggests that there is an effect or a difference. For this exercise, the alternative hypothesis \( H_a: \mu_1 \eq \mu_2 \) posits that there is a significant difference between the average scores of Strategy A and Strategy B. This hypothesis is what researchers aim to provide evidence for.
If data provides sufficient evidence against the null hypothesis, we reject it in favor of the alternative hypothesis. This would mean concluding that Strategy A and Strategy B do, in fact, result in different average scores.
Significance Level
The significance level, often represented by the Greek letter \( \alpha\), is a threshold set by the researcher to judge whether the obtained results are statistically significant. In simpler terms, it is the probability of rejecting the null hypothesis when it is actually true. The most common significance level is 5% (0.05).
In our exercise, this 5% significance level implies that we are willing to accept a 5% chance that any conclusion rejecting the null hypothesis might be incorrect. This helps us to control the likelihood of making a Type I error, which occurs when we incorrectly reject the true null hypothesis.
P-value
The P-value helps to quantify the evidence against the null hypothesis. Specifically, it represents the probability of obtaining test results at least as extreme as the ones observed, assuming the null hypothesis is true. A lower P-value indicates stronger evidence against the null hypothesis.
In this exercise, we are given a P-value of 0.18. To make a decision, we compare this P-value to our significance level of 0.05. Since 0.18 is greater than 0.05, we do not have enough evidence to reject the null hypothesis.
Therefore, we conclude that the observed difference in average scores between Strategy A and Strategy B is not significant. This leads us to decide that the correct answer is (B), as there is not sufficient evidence to suggest a significant difference between the two strategies.

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Most popular questions from this chapter

A pharmaceutical company claims that a medicine will produce a desired effect for a mean time of 58.4 minutes. A government researcher runs a hypothesis test of 40 patients and calculates a mean of \(\bar{x}=59.5\) with a standard deviation of \(s=8.3 .\) What is the \(P\) -value? (A) \(P\left(t>\frac{59.5-58.4}{8.3 / \sqrt{40}}\right)\) with \(d f=39\) (B) \(P\left(t>\frac{59.5-58.4}{8.3 / \sqrt{40}}\right)\) with \(d f=40\) (C) \(2 P\left(t>\frac{59.5-58.4}{8.3 / \sqrt{40}}\right)\) with \(d f=39\) (D) \(2 P\left(t>\frac{59.5-58.4}{8.3 / \sqrt{40}}\right)\) with \(d f=40\) (E) \(2 P\left(z>\frac{59.5-58.4}{8.3 / \sqrt{40}}\right)\)

Suppose (25,30) is a \(90 \%\) confidence interval estimate for a population mean \(\mu\). Which of the following is a true statement? (A) There is a 0.90 probability that \(\bar{x}\) is between 25 and \(30 .\) (B) Of the sample values, \(90 \%\) are between 25 and 30 . (C) There is a o.9o probability that \(\mu\) is between 25 and 30 . (D) If 100 random samples of the given size are picked and a \(90 \%\) confidence interval estimate is calculated from each, \(\mu\) will be in 90 of the resulting intervals. (E) If \(90 \%\) confidence intervals are calculated from all possible samples of the given size, \(\mu\) will be in \(90 \%\) of these intervals.

Most recent tests and calculations estimate at the \(95 \%\) confidence level that mitochondrial Eve, the maternal ancestor to all living humans, lived \(138,000 \pm 18,\) ooo years ago. What is meant by "95\% confidence" in this context? (A) A confidence interval of the true age of mitochondrial Eve has been calculated using \(z\) -scores of ±1.96 . (B) A confidence interval of the true age of mitochondrial Eve has been calculated using \(t\) -scores consistent with \(d f\) \(=n-1\) and tail probabilities of \(\pm 0.025 .\) (C) There is a 0.95 probability that mitochondrial Eve lived between 120,000 and 156,000 years ago. (D) If 20 random samples of data are obtained by this method and a \(95 \%\) confidence interval is calculated from each, the true age of mitochondrial Eve will be in 19 of these intervals. (E) Of all random samples of data obtained by this method, \(95 \%\) will yield intervals that capture the true age of mitochondrial Eve.

Suppose you do five independent tests of the form \(H_{0}: \mu=38\) versus \(H_{a}: \mu>38,\) each at the \(\alpha=0.01\) significance level. What is the probability of committing a Type I error and incorrectly rejecting a true null hypothesis with at least one of the five tests? (A) 0.01 (B) 0.049 (C) 0.05 (D) 0.226 (E) \(0.95^{1}\)

Does socioeconomic status relate to age at the time of HIV infection? For 274 high-income HIV-positive individuals, the average age of infection was 33.0 years with a standard deviation of 6.3 , while for 90 low-income individuals, the average age was 28.6 years with a standard deviation of 6.3 (The Lancet, October 22, 1994, page 1121). Find a \(90 \%\) confidence interval estimate for the difference in ages of all high- and low-income people at the time of HIV infection. (A) \(4.4 \pm 0.963\) (B) \(4.4 \pm 1.27\) (C) \(4.4 \pm 2.51\) (D) \(30.8 \pm 2.51\) (E) \(30.8 \pm 6.3\)

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