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To test the effect of a physical-fitness \- course on one's physical ability, the number of sit-ups that a person could do in 1 minute, both before and after the course, was recorded. Ten randomly selected participants scored as shown in the following table. Can you \- conclude that a significant amount of improvement took place? Use \(\alpha=0.01\) and assume normality.$$\begin{array}{lllllllllll}\text { Before } & 29 & 22 & 25 & 29 & 26 & 24 & 31 & 46 & 34 & 28 \\\\\text { After } & 30 & 26 & 25 & 35 & 33 & 36 & 32 & 54 & 50 & 43 \\\\\hline\end{array}$$a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

Short Answer

Expert verified
Using either the P-value or the classical approach, conclude whether the improvement is significant by comparing the statistic with the critical value or p-value with the significance level. The answer will depend on the exact calculations.

Step by step solution

01

Declare the Null Hypothesis and Alternate Hypothesis

The null hypothesis (\(H_0\)) should state that there is no significant difference in the average sit-ups before and after the course, thus \(H_0: \mu_D = 0\). The alternative hypothesis (\(H_1\)) should state that there is a significant increase in the average sit-ups after the course, thus \(H_1: \mu_D > 0\) where \(\mu_D\) is the mean difference of 'before' and 'after' measurements.
02

Calculate the Differences

Subtract the number of sit-ups before the course from the number of sit-ups after the course for each participant and calculate the mean difference. Let's denote this difference as \(d\), and as any observation can be written as 'before - after', negative values indicate improvement.
03

Calculate the t-statistic

The t-statistic is calculated as \(t= \frac{\bar{d}-\mu_D}{s_d/\sqrt{n}}\) where \(\bar{d}\) is the mean of the difference, \(\mu_D\) is the mean difference under the null hypothesis (which is 0 in this case), \(s_d\) is the standard deviation of the difference, and \(n\) is the number of pairs.
04

P-value Approach

For the P-value approach, compute the p-value using the t-distribution with \(n-1\) degrees of freedom. If this calculated p-value is less than the significance level (\(0.01\)), then reject the null hypothesis and conclude that significant improvement has taken place.
05

Classical Approach

For the classical approach, compare the calculated t-statistic to the critical t-value for \(n-1\) degrees of freedom and a significance level of \(0.01\). If the calculated t-value is greater than the critical value, reject the null hypothesis and conclude that significant improvement has taken place.

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Key Concepts

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

Null and Alternative Hypothesis
In hypothesis testing, the null hypothesis (H0) serves as the default assumption that there is no effect or no difference. In the context of our exercise, the null hypothesis posits that the physical-fitness course did not make a significant difference in the number of sit-ups a person can do in one minute. Formally, it is stated as H0: μD = 0, where μD represents the mean difference in sit-up counts before and after the course.

Contrarily, the alternative hypothesis (H1) suggests that an effect or difference does exist. For our exercise, the alternative hypothesis claims there is a significant increase in sit-up count post-course. This is denoted as H1: μD > 0. It is important to note that H1 is what we aim to provide evidence for through our test.

  • H0: No significant difference (μD = 0)
  • H1: Significant increase (μD > 0)
The ultimate goal of hypothesis testing is to determine whether to retain the null hypothesis or to support the alternative hypothesis based on the data.
T-test for Dependent Means
The t-test for dependent means, also known as the paired-samples t-test, is useful when comparing two related groups or conditions. In our exercise, we use this test to analyze the difference in sit-up counts for participants before and after undergoing the physical-fitness course. These are dependent samples because the same subjects are tested twice under different conditions.

To perform the test, we first calculate the difference for each pair of observations. Next, we compute the mean and standard deviation of these differences. The t-statistic is then calculated using the formula: t = (&bar;d - μD) / (sd / √n), where &bar;d is the average of the differences, μD is the hypothesized mean difference (0 in this case), sd is the standard deviation of differences, and n is the number of pairs. The calculated t-statistic allows us to determine how far the observed results are from what the null hypothesis predicts, relative to the variability in the data.
P-value Approach
The p-value represents the probability of obtaining the observed results, or something more extreme, if the null hypothesis is actually true. The lower the p-value, the greater the statistical significance of our observed differences.

In hypothesis testing, we compare the calculated p-value to a pre-determined significance level (α), which is the threshold for rejecting the null hypothesis. If the p-value is less than or equal to α, we reject the null hypothesis in favor of the alternative. Using the p-value approach, our exercise required calculating the p-value from the t-distribution with n-1 degrees of freedom. A p-value smaller than α = 0.01 indicates that the improvement in sit-up counts is considered statistically significant.
Significance Level
The significance level (α) is a critical concept in hypothesis testing, denoting the probability of making a Type I error, which is rejecting a true null hypothesis. The significance level is a boundary that helps us decide whether to reject H0 based on the p-value.

In our case, we used a significance level of α = 0.01, indicating we have a 1% risk of concluding that an improvement occurred when, in fact, it didn't (a false positive). The choice of α reflects our tolerance for risk. A small α reduces the chance of a Type I error but increases the risk of a Type II error, failing to detect an actual effect when one exists. Therefore, setting the significance level is an important step in hypothesis testing as it influences the interpretation of results and subsequent conclusions.

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Most popular questions from this chapter

Penfield and Perinton are two adjacent eastside suburbs of Rochester, New York. They have both always been considered on equal ground with respect to quality of life, housing, and education. Although many new housing developments are going up in Penfield, Perinton offers the added bonus of cheaper energy rates. It is thought that Perinton is able to have higher asking prices because of this bonus. To test this theory, random samples of real estate transactions were taken in each suburb during the week of October 17 \(2009 .\) Do the data support the theory for this time frame? Use \(\alpha=0.10\) $$\begin{array}{lc} \text { Pentield } & \text { Perinton } \\ \hline 195.700 & 154,900 \\\137,500 & 429,000 \\\117,000 & 272,000 \\\115,000 & 160,609 \\\176,000 & 265,000 \\\149,013 & 144,000 \\\130,000 & 152,000 \\ 266,490 & 390,000 \\\152,000 & 130,300 \\\262,765 & 149,900 \\\\\hline\end{array}$$

Determine the critical values that would be used for the following hypothesis tests (using the classical approach) about the difference between two means with population variances unknown. a. \(\quad H_{a}: \mu_{1}-\mu_{2} \neq 0, n_{1}=26, n_{2}=16, \alpha=0.05\) b. \(\quad H_{a}: \mu_{1}-\mu_{2}<0, n_{1}=36, n_{2}=27, \alpha=0.01\) c. \(\quad H_{a}: \mu_{1}-\mu_{2}>0, n_{1}=8, n_{2}=11, \alpha=0.10\) d. \(\quad H_{a}: \mu_{1}-\mu_{2} \neq 10, n_{1}=14, n_{2}=15, \alpha=0.05\)

Consumer Reports conducted a survey of 1000 adults concerning wearing bicycle helmets. One of the questions presented to 25 - to 54 -year olds was whether they wear a helmet most of the time while biking or cycling. This was further broken down to whether or not they had a child at home. Eighty-seven percent of the age group that had a child at home reported that they wear a helmet most of the time, while \(74 \%\) of those without a child at home reported they wear a helmet most of the time. If the sample size is 340 for both age groups, is the proportion of helmet use significantly greater when there is a child in the home, at the 0.01 level of significance?

The computer age has allowed teachers to use electronic tutorials to motivate their students to learn. Issues in Accounting Education published the results of a study that showed that an electronic tutorial, along with intentionally induced peer pressure, was effective in enhancing preclass preparations and in improving class attendance, test scores, and course evaluations when used by students studying tax accounting. Suppose a similar study is conducted at your school using an electronic study guide (ESG) as a tutor for students of accounting principles. For one course section, the students were required to use a new ESG computer program that generated and scored chapter review quizzes and practice exams, presented textbook chapter reviews, and tracked progress. Students could use the computer to build, take, and score their own simulated tests and review materials at their own pace before they took their formal in-class quizzes and exams composed of different questions. The same instructor taught the other course section, used the same textbook, and gave the same daily assignments, but he did not require the students to use the ESG. Identical tests were administered to both sections, and the mean scores of all tests and assignments at the end of the year were tabulated: $$\begin{array}{lccc}\text { Section } & n & \text { Mean Score } & \text { Std. Dev. } \\\\\hline \text { ESG }|1\rangle & 38 & 79.6 & 6.9 \\\\\text { No ESG }|2| & 36 & 72.8 & 7.6 \\\\\hline\end{array}$$ Do these results show that the mean scores of tests and assignments for students taking accounting principles with an ESG to help them are significantly greater than the mean scores of those not using an ESG? Use a 0.01 level of significance. a. Solve using the \(p\) -value approach.

A survey was conducted to determine the proportion of Democrats as well as Republicans who support a "get tough" policy in South America. The results of the survey were as follows: Democrats: \(n=250,\) number in support \(=120\) Republicans: \(n=200,\) number in support \(=105\) Construct the \(98 \%\) confidence interval for the difference between the proportions of support.

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