Question

The paper "Playing Active Video Games Increases Energy Expenditure in Children” (Pediatrics [2009]: \(534-539\) ) describes a study of the possible cardiovascular benefits of active video games. Mean heart rate for healthy boys ages 10 to 13 after walking on a treadmill at \(2.6 \mathrm{~km} /\) hour for 6 minutes is known to be 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 bpm and 15 bpm, respectively. Assume that the sample of boys is representative of boys ages 10 to 13 and that the distribution of heart rates after 15 minutes of Wii Bowling is approximately normal. 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 \(\alpha=0.01\). (Hint: See Example 12.12\()\)

Short answer

Without actual computation, the exact P-value and the conclusion regarding either to reject \(H_{0}\) or not cannot be provided here in the short answer.

Step-by-step solution

State the Hypotheses

The null hypothesis (\(H_{0}\)) and the alternate hypothesis (\(H_{1}\)) need to be formulated. For this problem: \n - Null hypothesis (\(H_{0}\)): The mean heart rate after playing Wii Bowling is the same as the known mean heart rate after walking, i.e., \(\mu = 98\)\n - Alternate hypothesis (\(H_{1}\)): The mean heart rate after playing Wii Bowling is different from the known mean heart rate after walking, i.e., \(\mu \neq 98\)

Calculate the Test Statistic

Since the standard deviation of the population is not known and the sample size is below 30, a t-statistic will be derived using the formula: \n \(t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}\)\n Where: \(\bar{x}\) is the sample mean (101bpm), \(\mu\) is the hypothesized population mean (98bpm), \(s\) is the sample standard deviation (15bpm), and \(n\) is the sample size (14). After substituting these values into the formula: \n \(t = \frac{101 - 98}{\frac{15}{\sqrt{14}}}\)

Determine the P-value

The P-value can be obtained by looking up the calculated t-statistic value in the t-distribution table or by using statistical software. For determining the P-value using the t-table, the degrees of freedom (df) need to be calculated first. The df is obtained by subtracting 1 from the sample size (df = n - 1 = 14 - 1 = 13). If the calculated P-value would be less than the given significance level (\(\alpha = 0.01\)), the null hypothesis would be rejected.

Conclude the Hypothesis Test

If the P-value obtained in the previous step is less than \(\alpha = 0.01\), reject the null hypothesis in favor of the alternative hypothesis, suggesting that the mean heart rate after playing Wii Bowling is significantly different from the known mean heart rate after walking. If the P-value is not less than \(\alpha\), fail to reject the null hypothesis, indicating that the means are not significantly different.

Key concepts

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. In other words, it tells us how spread out the numbers are from the average (mean) value. The standard deviation is crucial in hypothesis testing because it helps to determine the variability within a sample. In the context of the exercise,
the standard deviation for the heart rate of boys playing Wii Bowling was found to be 15 bpm. This value indicates how the individual heart rates of the boys deviate from the sample mean of 101 bpm. A larger standard deviation would suggest more variability in heart rates, while a smaller standard deviation would indicate that the heart rates are more closely clustered around the mean.

T-distribution

The t-distribution, also known as Student's t-distribution, is a probability distribution that arises when estimating the mean of a normally distributed population in situations where the sample size is small and the population standard deviation is unknown.

When applying this to our exercise, we use the t-distribution because the sample size is 14, which is less than 30, and the population standard deviation is not provided. The t-distribution is similar to the normal distribution but has heavier tails, meaning that it is more prone to producing values that fall far from the mean. As the sample size increases, the t-distribution approaches the normal distribution. The result of our calculated test statistic will be compared against a t-distribution to find out the P-value and determine whether our null hypothesis can be rejected.

P-value

The P-value is the probability of obtaining test results at least as extreme as the ones observed during the test, assuming that the null hypothesis is correct. A low P-value (<0.05 is a common threshold) indicates that the observed data is unlikely under the null hypothesis, leading to its rejection in favor of the alternate hypothesis.
In our exercise, the P-value tells us the likelihood of recording a sample mean heart rate of 101 bpm or more extreme, given that the true mean heart rate after walking is indeed 98 bpm. If this P-value is less than the significance level of 0.01, the difference in heart rates is considered statistically significant, and we would reject the null hypothesis.

Null Hypothesis

The null hypothesis ((H_0)) represents a statement of no effect or no difference and is the default assumption that there is no change or no association. In hypothesis testing, it serves as the claim that any observed effect is a result of sampling error.

In the study, the null hypothesis claims that the mean heart rate for boys playing Wii Bowling is the same as the mean heart rate after walking, specifically 98 bpm. It is essentially the skeptic's position, and the purpose of hypothesis testing is to determine whether the sample data provide enough evidence to cast doubt on this position.

Alternate Hypothesis

Conversely, the alternate hypothesis ((H_1) or (H_a)) represents the statement that there is an effect or a difference, in contrast to the null hypothesis. It conveys that observed effects from the sample data are due to true differences or changes rather than chance.
For the exercise in question, the alternate hypothesis posits that the mean heart rate for boys after playing Wii Bowling is different from the known mean heart rate after walking. This is what the researchers suspect and are attempting to provide evidence for through their hypothesis test.