/*! 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 14 In an effort to compare the aver... [FREE SOLUTION] | 91Ó°ÊÓ

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In an effort to compare the average swimming times for two swimmers, each swimmer was asked to swim freestyle for a distance of 100 yards at randomly selected times. The swimmers were thoroughly rested between laps and did not race against each other, so that each sample of times was an independent random sample. The times for each of 10 trials are shown for the two swimmers. \begin{tabular}{ll|ll} \multicolumn{2}{l|} { Swimmer 1} & \multicolumn{2}{c} { Swimmer 2 } \\ \hline 59.62 & 59.74 & 59.81 & 59.41 \\ 59.48 & 59.43 & 59.32 & 59.63 \\ 59.65 & 59.72 & 59.76 & 59.50 \\ 59.50 & 59.63 & 59.64 & 59.83 \\\ 60.01 & 59.68 & 59.86 & 59.51 \end{tabular} Suppose that swimmer 2 was last year's winner when the two swimmers raced. Does it appear that the average time for swimmer 2 is still faster than the average time for swimmer 1 in the 100 -yard freestyle? Find the approximate \(p\) -value for the test and interpret the results.

Short Answer

Expert verified
Answer: No, we cannot conclude that swimmer 2 is still faster than swimmer 1 in the 100-yard freestyle, as the p-value (0.292) is greater than the common significance level of 0.05, which means we do not have sufficient evidence to support the claim that swimmer 2 is faster.

Step by step solution

01

Calculate the mean and standard deviation for both swimmers

To begin, find the mean and standard deviation for the swimming times of both swimmers using the given data. These calculations can be done using a calculator or statistical software. For swimmer 1: Mean: 59.599 Standard deviation: 0.1795 For swimmer 2: Mean: 59.627 Standard deviation: 0.1649
02

Formulate null and alternative hypotheses

To test whether the average time for swimmer 2 is still faster than the average time for swimmer 1, we will use the following null and alternative hypotheses: Null hypothesis (H0): The average swimming time for both swimmers is the same, i.e., μ1 = μ2. Alternative hypothesis (H1): The average swimming time for swimmer 2 is faster than swimmer 1, i.e., μ1 > μ2.
03

Calculate the test statistic

Since we have two independent random samples and want to compare the means, we can use the two-sample t-test. The test statistic can be calculated using the following formula: t = (mean1 - mean2) / sqrt((std_dev1^2 / n1) + (std_dev2^2 / n2)) where n1 and n2 are the sample sizes for swimmer 1 and swimmer 2, respectively (in this case, n1 = n2 = 10). t = (59.599 - 59.627) / sqrt((0.1795^2 / 10) + (0.1649^2 / 10)) t ≈ -0.5541
04

Find the p-value and interpret the results

Now, we need to find the p-value for our test statistic. Since our alternative hypothesis is one-tailed (μ1 > μ2), we need to find the probability of getting a t-value greater than -0.5541 with 18 degrees of freedom (n1 + n2 - 2 = 18). Using a t-distribution table or statistical software, we find that the p-value is approximately 0.292. Since the p-value (0.292) is greater than the common significance level of 0.05, we fail to reject the null hypothesis, which means that we do not have sufficient evidence to conclude that the average time for swimmer 2 is still faster than the average time for swimmer 1 in the 100-yard freestyle. Keep in mind that this does not mean that swimmer 1 is faster than swimmer 2, but rather that the difference in their average swimming times is not statistically significant.

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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 concept in hypothesis testing. It serves as a starting assumption that there is no effect or no difference in the situation being analyzed. In the context of two-sample t-tests, the null hypothesis typically asserts that the means of the two populations are equal.
In our specific swimming times example, the null hypothesis \(H_0\) would state that there is no difference in the average times of Swimmer 1 and Swimmer 2. Mathematically, this is written as \(\mu_1 = \mu_2\), where \(\mu_1\) is the mean time of Swimmer 1 and \(\mu_2\) is the mean time of Swimmer 2.
Establishing the null hypothesis is crucial because it forms the basis for statistical testing. The aim is to determine whether there is enough evidence to reject this hypothesis, thus suggesting the presence of a significant difference.
alternative hypothesis
The alternative hypothesis, denoted as \(H_1\), contrasts the null hypothesis by proposing that a certain effect or difference does exist. In our exercise, the alternative hypothesis states that Swimmer 2 is still faster than Swimmer 1 on average.
This is articulated mathematically as \(\mu_1 > \mu_2\). It's essential to clearly define the direction of the hypothesis—whether it is one-tailed or two-tailed—because this affects the interpretation of the test results.
  • One-Tailed Test: Tests if one mean is greater than the other. In this case, Swimmer 1's time is hypothesized to be consistently longer than Swimmer 2's.
  • Two-Tailed Test: Tests for any difference in means without specifying the direction.
In our case, it's a one-tailed test since we are specifically interested in whether Swimmer 2 keeps a faster average time.
The outcome of our test will either provide evidence to support accepting the alternative hypothesis or it will not provide sufficient evidence, in which case we do not reject the null hypothesis.
p-value
The p-value is a crucial part of hypothesis testing and serves as a measure of the strength of evidence against the null hypothesis. It quantifies the probability of obtaining a test statistic that is as extreme as, or more extreme than, the observed result, assuming that the null hypothesis is true.
In simpler terms, a low p-value indicates that the observed data is less likely under the null hypothesis, thereby providing stronger evidence against it. Conversely, a high p-value suggests that the observed data is more consistent with the null hypothesis.
  • Thresholds: Common significance levels are 0.05 or 0.01, which dictate when to reject the null hypothesis.
In our swimming example, the p-value was found to be approximately 0.292. Since this value is greater than 0.05, we do not have sufficient evidence to reject the null hypothesis. Rather than suggesting Swimmer 1 is faster, this outcome implies that the data does not statistically demonstrate Swimmer 2's continued superiority.
Understanding p-values helps in making informed decisions about hypotheses in the face of uncertainty.
standard deviation
Standard deviation is a key statistical measure that represents the dispersion or variability within a set of data. It quantifies how much the individual data points differ from the mean of the data set.
Smaller standard deviations indicate that the data points tend to be close to the mean, reflecting more consistency, while larger standard deviations suggest greater variability and a wider range of values.
In the swimming times scenario:
  • Swimmer 1 has a standard deviation of 0.1795.
  • Swimmer 2 has a standard deviation of 0.1649.
Both swimmers have relatively small standard deviations, meaning their lap times are consistently close to their respective means.
The standard deviation is essential when performing a two-sample t-test because it helps in measuring the amount of variation within each sample. This, in turn, affects the test statistic calculation, which plays a vital role in comparing the two means.
The smaller the standard deviation, the more reliable our estimates of the mean, leading to more powerful tests.

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