Question

The effect of alcohol consumption on the body appears to be much greater at high altitudes than at sea level. To test this theory, a scientist randomly selects 12 subjects and randomly divides them into two groups of six each. One group is put into a chamber that simulates conditions at an altitude of 12,000 feet, and each subject ingests a drink containing 100 cubic centimeters (cc) of alcohol. The second group receives the same drink in a chamber that simulates conditions at sea level. After 2 hours, the amount of alcohol in the blood (grams per 100 cc ) for each subject is measured. The data are shown in the table. Do the data provide sufficient evidence to support the theory that average amount of alcohol in the blood after 2 hours is greater at high altitudes? $$ \begin{array}{cc} \text { Sea Level } & 12,000 \text { Feet } \\ \hline .07 & .13 \\ .10 & .17 \\ .09 & .15 \\ .12 & .14 \\ .09 & .10 \\ .13 & .14 \end{array} $$

Short answer

Answer: Yes, based on the performed hypothesis test, there is sufficient evidence to support the theory that the average amount of alcohol in the blood after 2 hours is greater at high altitudes than at sea level.

Step-by-step solution

State the null and alternative hypotheses

The null hypothesis (H0) states that there is no difference in the average amount of alcohol in the blood after 2 hours between the sea level group and the 12,000 feet group. The alternative hypothesis (H1) states that there is a difference, with the average amount of alcohol in the blood after 2 hours being greater at 12,000 feet. H0: μ1 - μ2 = 0 H1: μ1 - μ2 ≠ 0 Where μ1 is the average amount of alcohol in the blood at sea level and μ2 is the average amount of alcohol in the blood at 12,000 feet.

Calculate the sample mean and sample standard deviation

Calculate the sample means (x̄1 and x̄2) and the sample standard deviations (s1 and s2) for each group. For the sea level group: x̄1 = (0.07 + 0.10 + 0.09 + 0.12 + 0.09 + 0.13) / 6 = 0.1 s1 = sqrt(Σ(xi - x̄1)^2 / (n1 - 1)) = sqrt(0.00073) ≈ 0.027 For the 12,000 feet group: x̄2 = (0.13 + 0.17 + 0.15 + 0.14 + 0.10 + 0.14) / 6 = 0.13833 s2 = sqrt(Σ(xi - x̄2)^2 / (n2 - 1)) = sqrt(0.000798) ≈ 0.028

Calculate the test statistic

We'll perform a two-sample t-test to compare the means of the two groups. Calculate the test statistic (t) using the formula: t = (x̄1 - x̄2) / sqrt((s1^2 / n1) + (s2^2 / n2)) t = (0.1 - 0.13833) / sqrt((0.027^2 / 6) + (0.028^2 / 6)) ≈ -3.153

Determine the critical value and the p-value

For a two-tailed test with α = 0.05, and degrees of freedom (df) = n1 + n2 - 2 = 6 + 6 - 2 = 10, the critical value is approximately ± 2.228. We can also calculate the p-value using the t-distribution table or a calculator. For our test statistic, the p-value is approximately 0.011.

Make a decision

Since the calculated t statistic (-3.153) is less than the critical value (-2.228), we reject the null hypothesis (H0) in favor of the alternative hypothesis (H1). Alternatively, the p-value of 0.011 is less than our chosen significance level of 0.05, which also leads us to reject the null hypothesis. Based on this statistical test, there is sufficient evidence to support the theory that the average amount of alcohol in the blood after 2 hours is greater at high altitudes than at sea level.

Key concepts

Two-Sample T-Test

A two-sample t-test is a statistical method used to determine whether there is a significant difference between the means of two independent groups. For example, in our exercise, we wanted to know if the average level of alcohol in the blood after two hours is different for people staying at sea level compared to those at a high altitude of 12,000 feet.
The steps involved in conducting a two-sample t-test include calculating the means and standard deviations for both groups and then determining the test statistic, often denoted as \( t \).
When we set up a two-sample t-test, our main goal is to assess if the observed differences in sample means could have occurred by random chance, or if they are statistically significant. It helps in making informed conclusions about the population from which the samples were drawn.

Null and Alternative Hypotheses

In any hypothesis test, we start with two opposing hypotheses: the null hypothesis \((H_0)\) and the alternative hypothesis \((H_1)\).
The null hypothesis \((H_0)\) represents a statement of no effect or no difference. In our exercise, it was stated as: "There is no difference in the average amount of alcohol in the blood between the sea-level group and the 12,000-feet group." Mathematically, this can be expressed as \( \mu_1 - \mu_2 = 0 \).
The alternative hypothesis \((H_1)\), on the other hand, states what we suspect might be true - a difference exists. For this exercise, it was suggested that the average amount of alcohol in the blood is greater at 12,000 feet compared to sea level, symbolically written as \( \mu_1 - \mu_2 eq 0 \).
The testing procedure revolves around the 'acceptance' or 'rejection' of these hypotheses based on the data.

P-Value

The p-value is a critical concept in hypothesis testing that helps determine the strength of the results. It represents the probability of obtaining test results at least as extreme as the observed results, assuming that the null hypothesis is true.
In simpler terms, a smaller p-value indicates that the evidence against the null hypothesis is stronger. For instance, in our exercise, the calculated p-value was approximately 0.011.
This means that there is only a 1.1% chance that the observed difference in blood alcohol levels (or a more extreme difference) could occur if there were actually no true difference between the two groups.

• If the p-value is less than the chosen significance level (commonly 0.05 or 5%), we reject the null hypothesis.
• If the p-value is greater, we do not reject the null hypothesis.

In our case, the p-value was less than 0.05, leading to the rejection of the null hypothesis.

Critical Value

Critical value is a point on the test distribution that is compared to the test statistic to decide whether to reject the null hypothesis.
For a two-tailed test, the critical values divide the t-distribution into two tails. Each tail represents the level of significance, \( \alpha \), which is essentially the probability of rejecting the null hypothesis when it is true (a type I error).
In our case, with a significance level of 0.05 and degrees of freedom involving both groups (\( df = n_1 + n_2 - 2 = 10 \)), the critical value was approximately ±2.228.
If the calculated test statistic falls beyond this critical value, it suggests that the null hypothesis should be rejected.
For our exercise, the test statistic (-3.153) was more extreme than the negative critical value (-2.228), indicating that the effect of alcohol consumption is indeed greater at higher altitudes.