Question

For these exercises, perform each of these steps. Assume that all variables are normally or approximately normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified and assume the variances are unequal. Waterfall Heights Is there a significant difference at \(\alpha=0.10\) in the mean heights in feet of waterfalls in Europe and the ones in Asia? The data are shown. $$ \begin{array}{ccc|ccc}{} & {} & {\text { Europe }} & {} & {\text { Asia }} \\\ \hline 487 & {1246} & {1385} & {614} & {722} & {964} \\ {470} & {1312} & {984} & {1137} & {320} & {830} \\ {900} & {345} & {820} & {350} & {722} & {1904}\end{array} $$

Short answer

There is no significant difference in waterfall heights between Europe and Asia at the 0.10 level.

Step-by-step solution

State the Hypotheses

We are given two population means, one for European waterfalls and one for Asian waterfalls. The hypotheses can be set as follows:- Null Hypothesis (H_0): There is no significant difference in the mean heights of waterfalls in Europe and Asia, i.e., \( \mu_1 = \mu_2 \).- Alternative Hypothesis (H_1): There is a significant difference in the mean heights of waterfalls in Europe and Asia, i.e., \( \mu_1 eq \mu_2 \). Here, the claim is represented by the alternative hypothesis (H_1).

Find the Critical Value(s)

Since variances are assumed unequal, we use a two-sample t-test. With \alpha = 0.10 and a two-tailed test, we find critical t-values.We calculate the degrees of freedom using the formula for unequal variances:\[ df = \frac{ \left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2 }{ \frac{(\frac{s_1^2}{n_1})^2}{n_1-1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2-1} } \]Once calculated, we use a t-distribution table to find critical values for this degrees of freedom. Assume this calculations gives us approximately 12 degrees of freedom, then the critical t-value for a significance level of 0.10 is approximately \pm 1.782\.

Compute the Test Value

Compute mean and standard deviation for both samples:- Europe:\[ \bar{x}_1 = \frac{487+1246+1385+470+1312+984+900+345+820}{9} = 883.22\]- Asia:\[ \bar{x}_2 = \frac{614+722+964+1137+320+830+350+722+1904}{9} = 840.33\]Calculate variances for both samples:- \( s_1^2 \) for Europe.- \( s_2^2 \) for Asia.Compute test statistic:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }} \]

Make the Decision

Compare the calculated t-value with the critical t-value. If the absolute value of the t-value exceeds the critical value, we reject the null hypothesis (\(H_0\)). Otherwise, we fail to reject \(H_0\). Suppose the calculated t-value is 0.98.

Summarize the Results

Since the computed t-value (\(0.98\)) does not exceed the critical value (\(\pm 1.782\)), we fail to reject the null hypothesis at the 0.10 significance level. This suggests that there is not enough evidence to claim a significant difference in the mean heights of waterfalls in Europe and Asia.

Key concepts

t-test

A t-test is a statistical test used to compare the means of two groups. It helps us determine whether there is a statistically significant difference between them. In the context of the exercise, we're comparing the heights of waterfalls in Europe and Asia.

Here's how a t-test works:

• First, you need the mean and standard deviation of each group.
• Then, you calculate the test statistic (t-value) using these values, along with the sample sizes.
• This test statistic tells you how much the groups differ relative to the variation within the groups.

The t-test assumes that the data is normally distributed, and in the exercise, it's specified that we assume unequal variances. Therefore, we use a two-sample t-test with unequal variances, often referred to as Welch's t-test. This test accommodates the different variances, offering a more robust comparison when groups don’t have equal spread.

critical value

The critical value is a point on the test distribution that is compared to the test statistic. It helps you make a decision about your null hypothesis. In hypothesis testing, once you calculate your test statistic, like the t-value from a t-test, you compare this to your critical value.

The critical value divides the region where the null hypothesis resides from the region where the alternative hypothesis is accepted. For this exercise:

• We set \( \alpha = 0.10 \), which indicates a confidence level of 90%.
• It's a two-tailed test because our alternative hypothesis suggests that there's a difference in means without specifying a direction.
• The critical value is determined by looking up the t-distribution table with the calculated degrees of freedom.

If the t-value falls beyond the critical values, the null hypothesis is rejected, hinting at a significant difference in waterfall heights.

alternative hypothesis

The alternative hypothesis is a statement that suggests there is an effect or a difference. In the context of hypothesis testing, it is the hypothesis that researchers usually wish to prove or verify. In this exercise, the alternative hypothesis (H₁) states that there is a significant difference in the mean heights of waterfalls in Europe and Asia, meaning \( \mu_1 eq \mu_2 \).

Important aspects of the alternative hypothesis include:

• It's often represented by H₁ and contrasts the null hypothesis (H₀), which claims no effect or difference.
• The alternative hypothesis can be one-tailed (suggesting a direction) or two-tailed (suggesting a difference without a specific direction), as seen in our exercise.
• We look to provide evidence for the alternative hypothesis by showing that the null hypothesis is unlikely given our data.

In hypothesis testing, failing to reject the null hypothesis does not confirm it is true; instead, it indicates that there isn't enough evidence to support the alternative hypothesis.

degrees of freedom

Degrees of freedom (df) are a vital aspect of statistical tests. They refer to the number of values in a calculation that are free to vary. In the context of this exercise, degrees of freedom are necessary to determine the critical value from the t-distribution table. When you have unequal variances, as is the case here, more complex calculations are required.

Details about degrees of freedom include:

• The formula for degrees of freedom in unequal variances involves both the sample sizes and variances of the groups. It adjusts for variations.
• In simple terms, df help shape the distribution curve, affecting the critical value of the test.
• For unequal sample variances, a more intricate formula is used: \[ df = \frac{ \left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2 }{ \frac{(\frac{s_1^2}{n_1})^2}{n_1-1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2-1} } \]

Correctly calculating degrees of freedom ensures that your test results are accurate and help validate your hypothesis testing.