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The following is based on information taken from Winter Wind Studies in Rocky Mountain National Park, by D. E. Glidden (Rocky Mountain Nature Association). At five weather stations on Trail Ridge Road in Rocky Mountain National Park, the peak wind gusts (in miles per hour) for January and April are recorded below. \(\begin{array}{l|ccccc} \hline \text { Weather Station } & 1 & 2 & 3 & 4 & 5 \\ \hline \text { January } & 139 & 122 & 126 & 64 & 78 \\ \hline \text { April } & 104 & 113 & 100 & 88 & 61 \\ \hline \end{array}\) Does this information indicate that the peak wind gusts are higher in January than in April? Use \(\alpha=0.01\).

Short Answer

Expert verified
The data indicates that the peak wind gusts are significantly higher in January than in April.

Step by step solution

01

Formulate Hypotheses

Define the null hypothesis H_0, which states that there is no difference in the mean peak wind gusts between January and April. Define the alternative hypothesis H_a, which states that the mean peak wind gusts in January are greater than in April: \(H_0: \mu_J = \mu_A\) (null hypothesis) \(H_a: \mu_J > \mu_A\) (alternative hypothesis)
02

Calculate Mean and Standard Deviation

Calculate the means and standard deviations for the January and April wind speeds. For January: Mean: \(\bar{x}_J = \frac{139 + 122 + 126 + 64 + 78}{5}\)Standard Deviation: Use similar to mean, using the formula \(s_J = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\)For April: Mean: \(\bar{x}_A = \frac{104 + 113 + 100 + 88 + 61}{5}\)Standard Deviation: \(s_A\)
03

Perform Two-Sample t-Test

Use the Two-Sample t-Test formula for unequal variances: \[t = \frac{\bar{x}_J - \bar{x}_A}{\sqrt{\frac{s_J^2}{n_J} + \frac{s_A^2}{n_A}}}\]where \(n_J = n_A = 5\). Calculate the test statistic \(t\).
04

Determine Degrees of Freedom and Critical Value

Given the significance level \(\alpha = 0.01\), use degrees of freedom calculated using the formula: \[ df = \left( \frac{\left(\frac{s_J^2}{n_J} + \frac{s_A^2}{n_A}\right)^2}{\frac{(\frac{s_J^2}{n_J})^2}{n_J - 1} + \frac{(\frac{s_A^2}{n_A})^2}{n_A - 1}} \right) \]Find the critical value from the t-distribution table.
05

Make a Decision

If the calculated t-value exceeds the critical t-value from the t-table, reject the null hypothesis. Otherwise, do not reject it.

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Key Concepts

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

Two-Sample T-Test
A two-sample t-test is a statistical method used to determine if there is a significant difference between the means of two groups. Imagine you have two sets of data, like the peak wind gusts in January and April, and want to know if one set is statistically higher than the other. This test is handy for comparing means when dealing with continuous data from two independent groups.
  • The test assumes that both groups are normally distributed.
  • It can be used with either equal or unequal variances.
When using a two-sample t-test, it's essential to evaluate whether the variances (a measure of how data points differ from the mean) are equal or not. This will dictate which version of the test to use. For our weather data, since we assume unequal variances, we use the formula for the two-sample t-test with unequal variances to calculate our test statistic.
Once the calculations are complete, the test statistic is compared against a critical value from a t-distribution table to help us make a decision regarding our hypotheses.
Null Hypothesis
In hypothesis testing, the null hypothesis (H_0) is a statement of no effect or no difference. It's the hypothesis that we initially assume is true. It acts as a default position, asserting that any observed effect is due to random chance.
For the exercise at hand, the null hypothesis is: \( H_0: \mu_J = \mu_A \). This hypothesis postulates that the mean peak wind gusts in January are equal to those in April.
  • The null hypothesis often serves as a starting point for statistical testing.
  • Rejecting the null hypothesis suggests that there is sufficient evidence to support an alternative claim.
In practice, we perform hypothesis testing to determine whether there is enough statistical evidence in a sample to infer that the null hypothesis does not hold true in the population.
Alternative Hypothesis
The alternative hypothesis (H_a) is contrary to the null hypothesis and represents what we aim to support through our statistical analysis. It claims that there is an effect or a difference, often driven by the original question or study objective.
For our problem, the alternative hypothesis is:\( H_a: \mu_J > \mu_A \). This means we suspect that January's mean peak wind gusts are greater than those in April.
  • The alternative hypothesis is what researchers usually hope to affirm.
  • It is formulated to represent the effect or change being tested against the null hypothesis.
Testing against the alternative hypothesis helps to provide evidence of a significant difference or effect that might not be due to mere chance.
Significance Level
The significance level, denoted by \( \alpha \), is a threshold set by researchers to determine when to reject the null hypothesis. It quantifies the risk we are willing to take of rejecting a true null hypothesis (a type I error). Common significance levels are 0.05, 0.01, and 0.10.
In this exercise, \( \alpha = 0.01 \), meaning there is a 1% risk of concluding that there is a difference when, in fact, there is none. A lower significance level like 0.01 implies a stricter criterion for hypothesis testing.
  • A small \( \alpha \) reduces the likelihood of falsely detecting an effect that doesn't exist.
  • It's important to balance \( \alpha \) to avoid type I errors while still being able to detect true effects.
When the calculated test statistic falls beyond the critical value at this level of significance, we reject the null hypothesis in favor of the alternative, indicating enough evidence of a significant difference.

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Most popular questions from this chapter

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