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The paper "Playing Active Video Games Increases Energy Expenditure in Children" (Pediatrics [2009]: 534-539) describes an interesting investigation of the possible cardiovascular benefits of active video games. Mean heart rate for healthy boys age 10 to 13 after walking on a treadmill at \(2.6 \mathrm{~km} /\) hour for 6 minutes is 98 beats per minute (bpm). For each of 14 boys, heart rate was measured after 15 minutes of playing Wii Bowling. The resulting sample mean and standard deviation were \(101 \mathrm{bpm}\) and \(15 \mathrm{bpm}\), respectively. For purposes of this exercise, assume that it is reasonable to regard the sample of boys as representative of boys age 10 to 13 and that the distribution of heart rates after 15 minutes of Wii Bowling is approximately normal. a. Does the sample provide convincing evidence that the mean heart rate after 15 minutes of Wii Bowling is different from the known mean heart rate after 6 minutes walking on the treadmill? Carry out a hypothesis test using \(\alpha=.01\). b. The known resting mean heart rate for boys in this age group is \(66 \mathrm{bpm}\). Is there convincing evidence that the mean heart rate after Wii Bowling for 15 minutes is higher than the known mean resting heart rate for boys of this age? Use \(\alpha=.01\).

Step by step solution

01

Set Up the Hypotheses for the First Test

The null hypothesis (\(H_0\)) is that the mean heart rate after Wii Bowling is equal to the known mean heart rate after walking on a treadmill, i.e., \(H_0: \mu = 98\). The alternative hypothesis (\(H_a\)) is that the two means are not equal, i.e., \(H_a: \mu \neq 98\).

02

Perform the First Test

Conduct a t-test using the formula for t-statistic: \[t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}\] where \(\bar{x}\) is the sample mean, \(\mu_0\) is the population mean under the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size. Here, \(\bar{x}=101\), \(\mu_0=98\), \(s=15\), and \(n=14\). Calculate the t-statistic from these values and compare it to the critical t-value for a two-tailed test with \(\alpha=.01\) and \(df = n - 1\). If the calculated t-statistic is greater than the critical t-value, reject the null hypothesis.

03

Set Up the Hypotheses for the Second Test

The null hypothesis (\(H_0\)) for this scenario is that the mean heart rate after Wii Bowling is equal to the known mean resting heart rate, i.e., \(H_0: \mu = 66\). The alternative hypothesis (\(H_a\)) is that the mean heart rate after Wii Bowling is greater than the known mean resting heart rate, i.e., \(H_a: \mu > 66\).

04

Perform the Second Test

Again, perform a t-test. Here, \(\bar{x}=101\), \(\mu_0=66\), \(s=15\), and \(n=14\). Calculate the t-statistic from these values and compare it to the critical t-value for a one-tailed test with \(\alpha=.01\) and \(df = n - 1\). If the calculated t-statistic is greater than the critical t-value, reject the null hypothesis.

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Key Concepts

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

t-test

The t-test is a statistical procedure used to determine whether there is a significant difference between the means of two groups or whether a sample mean significantly differs from a known population mean. It is essential in comparing averages to find out if the observed results could happen just by chance or if they imply a real effect or difference.

When conducting a t-test, the goal is to compute a t-statistic, which measures how far the sample mean is from the population mean, normalized by the variation in the sample data. Depending on the hypothesis and data, different forms of the t-test are applied, such as one-sample, independent two-sample, or paired sample t-tests. In the context of the original exercise, a one-sample t-test was employed because the research is comparing a single sample mean to a known population mean.

sample mean

The sample mean, denoted as \(\bar{x}\), is the average value of a sample and serves as an estimate of the population mean. It's calculated by summing all the observations in the sample and dividing by the number of observations. In hypothesis testing, the sample mean is compared with a population mean to assess whether any observed differences are statistically significant. In the example exercise, the sample mean heart rate of 101 bpm after Wii Bowling is compared with the known population means.

standard deviation

Standard deviation, represented as \(s\), is a measure of the dispersion or spread in a set of values. A high standard deviation implies that the values in a dataset are spread out over a broader range of values, while a low standard deviation indicates that they are more clustered around the mean.

In hypothesis testing, standard deviation is used in the calculation of the t-statistic to normalize the difference between the sample mean and the population mean; it reflects how much variability there is from the mean within the data set. In the student's exercise, a standard deviation of 15 bpm means there's considerable variability in the heart rates after playing Wii Bowling.

null hypothesis

The null hypothesis, denoted as \(H_0\), is a statement of no effect or no difference, and it serves as the starting assumption for statistical significance testing. It is the hypothesis that the researcher seeks to test and possibly reject in favor of an alternative hypothesis based on sample data.

In the given exercise, the null hypothesis for the first test is that the mean heart rate after Wii Bowling \(H_0: \mu = 98\) bpm, is the same as the mean heart rate after walking on a treadmill.

alternative hypothesis

Contrary to the null hypothesis, the alternative hypothesis, denoted as \(H_a\) or \(H_1\), suggests that there is an effect or a difference, and it represents what the researcher wants to prove. It's the statement that will be concluded if the null hypothesis is rejected.

For instance, the alternative hypothesis in the first part of our exercise scenario was \(H_a: \mu eq 98\), stating that the mean heart rate after Wii Bowling is not equal to 98 bpm, implying that it could be either higher or lower.

statistical significance

Statistical significance indicates whether the result of a test is unlikely to have occurred by random chance alone. A result is said to be statistically significant if the observed p-value is less than the predetermined significance level \(\alpha\), which is typically set at 0.05 or 0.01.

In our original problem, the significance level for the tests was set at \(\alpha = .01\), meaning there is only a 1% chance that the observed differences in heart rate could have occurred if the null hypothesis were true. A statistically significant result would lead to the rejection of the null hypothesis.

critical value

The critical value is a cutoff point on the scale of the test statistic that determines whether to reject the null hypothesis. For a given significance level \(\alpha\), it is the value that divides what are considered common outcomes from rare ones under the assumption that the null hypothesis is true.

In hypothesis testing, if the test statistic exceeds the critical value, the result is declared statistically significant, and the null hypothesis is rejected. Depending on whether the test is one-tailed or two-tailed, as well as the degrees of freedom, different critical values are used. For the exercise at hand, comparing the t-statistic to the critical value corresponding to a two-tailed test (first scenario) and a one-tailed test (second scenario) with \(\alpha=.01\) determines the outcome of the hypothesis tests.