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The effect of alcohol consumption on the body appears to be much greater at higher altitudes. To test this theory, a scientist randomly selected 12 subjects and divided them into two groups of 6 each. One group was transported to an altitude of 12,000 feet, and each member in the group ingested 100 cubic centimeters \(\left(\mathrm{cm}^{3}\right)\) of alcohol. The members of the second group were taken to sea level and given the same amount of alcohol. After 2 hours, the amount of alcohol in the blood of each subject was measured (measurements in grams/100 \(\mathrm{cm}^{3}\) ). The data are given in the following table. Is there sufficient evidence to indicate that retention of alcohol is greater at 12,000 feet than at sea level? Test at the \(\alpha=.10\) level of significance. $$\begin{array}{cc} \hline \text { Sea Level } & \begin{array}{c} 12,000 \\ \text { feet } \end{array} \\ \hline .07 & .13 \\ .10 & .17 \\ .09 & .15 \\ .12 & .14 \\ .09 & .10 \\ .13 & .14 \\ \hline \end{array}$$

Short Answer

Expert verified
There is sufficient evidence to indicate that alcohol retention is greater at 12,000 feet than at sea level at \(\alpha = 0.10\).

Step by step solution

01

Define Hypotheses

To test the hypothesis, first define the null hypothesis \(H_0\) and the alternative hypothesis \(H_a\). The null hypothesis \(H_0\) will be that the mean blood alcohol retention at 12,000 feet \((\mu_1)\) is equal to that at sea level \((\mu_2)\). The alternative hypothesis \(H_a\) is that the mean blood alcohol retention at 12,000 feet is greater than that at sea level. In symbols, \(H_0: \mu_1 = \mu_2\) and \(H_a: \mu_1 > \mu_2\).
02

Determine Significance Level

The level of significance \(\alpha\) is given as 0.10. This means you will reject the null hypothesis if the p-value is less than 0.10.
03

Calculate Sample Means

Calculate the sample mean for both groups. The sea level group mean \(\bar{x}_2 = \frac{0.07 + 0.10 + 0.09 + 0.12 + 0.09 + 0.13}{6} = 0.1\). The 12,000 feet group mean \(\bar{x}_1 = \frac{0.13 + 0.17 + 0.15 + 0.14 + 0.10 + 0.14}{6} = 0.1383\).
04

Calculate Sample Variances

Calculate the sample variance for both groups. For sea level, \(s_2^2 = \frac{(0.07-0.1)^2 + (0.10-0.1)^2 + (0.09-0.1)^2 + (0.12-0.1)^2 + (0.09-0.1)^2 + (0.13-0.1)^2}{5} = 0.00064\). For 12,000 feet, \(s_1^2 = \frac{(0.13-0.1383)^2 + (0.17-0.1383)^2 + (0.15-0.1383)^2 + (0.14-0.1383)^2 + (0.10-0.1383)^2 + (0.14-0.1383)^2}{5} = 0.0009867\).
05

Calculate Test Statistic

Compute the test statistic using the formula for a two-sample t-test: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] where \(n_1 = n_2 = 6\). Substituting the values: \[ t = \frac{0.1383 - 0.1}{\sqrt{\frac{0.0009867}{6} + \frac{0.00064}{6}}} = \frac{0.0383}{0.0153} = 2.503 \]
06

Determine Critical t-Value

For a one-tailed test with \(\alpha = 0.10\) and degrees of freedom \(df = n_1 + n_2 - 2 = 10\), find the critical t-value from the t-distribution table, which is approximately 1.372.
07

Compare Test Statistic to Critical Value

Since the calculated t-value \(2.503\) is greater than the critical t-value \(1.372\), we reject the null hypothesis.

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Key Concepts

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

Null Hypothesis
A null hypothesis acts as a starting point in your statistical investigation. It is essentially a statement suggesting no effect or no difference. Here, the null hypothesis, denoted as \( H_0 \), assumes that there is no difference in blood alcohol retention between the two altitude levels. This means that the average alcohol retention in the bloodstream at 12,000 feet is equal to that at sea level. Formally, this can be expressed like so: \( H_0: \mu_1 = \mu_2 \), indicating both group means are the same.
It serves as the hypothesis your test aims to challenge. The null hypothesis is kept as the default assumption until evidence suggests otherwise. In hypothesis testing, you either reject the null hypothesis or fail to reject it, based on the data analyzed. By establishing \( H_0 \), researchers like the scientist in the exercise, set a clear framework to determine if the altitude truly affects alcohol retention sufficiently enough to infer it statistically.
Significance Level
The significance level, often denoted by \( \alpha \), is a crucial threshold in hypothesis testing. It represents the probability of rejecting the null hypothesis when it is actually true. A common choice for \( \alpha \) is 0.05, but in our scenario, it is set at 0.10. This indicates a 10% risk of mistakenly claiming a difference when none exists.
Choosing \( \alpha \) involves balancing risk and confidence in your claims. Lower values (like 0.01) provide more confidence but require stronger evidence to reject the null hypothesis. However, choosing a higher \( \alpha \) value (like 0.10) makes the test more sensitive and increases the risk of a type I error, where a true null hypothesis is incorrectly rejected.
Thus, when conducting a test at \( \alpha = 0.10 \), we become prepared to reject \( H_0 \) only if the test statistic falls into the critical region determined by this significance level.
Two-Sample t-Test
The two-sample t-test is a method used to determine whether there is a significant difference between the means of two independent samples. It's suitable when analyzing a small sample (fewer than 30), like the one in the exercise.- **Purpose and Use:** This statistical measure helps judge if the means of two populations are different, under the assumption that data follows a normal distribution. In the current exercise, it examines whether the blood alcohol retention at 12,000 feet is statistically greater than at sea level.
- **Calculation Insight:** Known by its specific formula, the two-sample t-test statistic allows comparison of sample means while accounting for their variance. The formula used here is: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] where \( \bar{x}_1 \) and \( \bar{x}_2 \) are the sample means, \( s_1^2 \) and \( s_2^2 \) are variances, and \( n_1 \), \( n_2 \) are the sizes of each group. By plugging in the calculated values, you can obtain the t-statistic, which indicates how far apart the sample means are, measured by the variability of both samples.
- **Decision Making:** Once you have the t-value, it’s compared with the critical t-value found in statistical tables. If your computed t-value exceeds the critical t-value (considering degrees of freedom and significance level), you conclude there is a significant difference between the two means.
In summary, a two-sample t-test is essential in validating claims of difference in experimental data, like the alcohol retention under different altitudes.

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Most popular questions from this chapter

Suppose that independent random samples of sizes \(n_{1}\) and \(n_{2}\) are to be selected from normal populations with means \(\mu_{1}\) and \(\mu_{2},\) respectively, and common variance \(\sigma^{2} .\) For testing \(H_{0}: \mu_{1}=\mu_{2}\) versus \(H_{a}: \mu_{1}-\mu_{2}>0\left(\sigma^{2} \text { unknown }\right)\), show that the likelihood ratio test reduces to the two ample \(t\) test presented in Section \(10.8 .\)

Suppose that \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denote a random sample from the probability density function given by $$f\left(y | \theta_{1}, \theta_{2}\right)=\left\\{\begin{array}{ll} \left(\frac{1}{\theta_{1}}\right) e^{-\left(y-\theta_{2}\right) / \theta_{1}}, & y > \theta_{2} \\ 0, & \text { elsewhere } \end{array}\right.$$ Find the likelihood ratio test for testing \(H_{0}: \theta_{1}=\theta_{1,0}\) versus \(H_{a}: \theta_{1} > \theta_{1,0},\) with \(\theta_{2}\) unknown.

The stability of measurements of the characteristics of a manufactured product is important in maintaining product quality. In fact, it is sometimes better to obtain small variation in the measured value of some important characteristic of a product and have the process mean slightly off target than to get wide variation with a mean value that perfectly fits requirements. The latter situation may produce a higher percentage of defective product than the former. A manufacturer of light bulbs suspected that one of his production lines was producing bulbs with a high variation in length of life. To test this theory, he compared the lengths of life of \(n=50\) bulbs randomly sampled from the suspect line and \(n=50\) from a line that seemed to be in control. The sample means and variances for the two samples were as shown in the following table. $$\begin{array}{ll} \hline \text { Suspect Line } & \text { Line in Control } \\ \hline \bar{y}_{1}=1,520 & \bar{y}_{2}=1,476 \\ s_{1}^{2}=92,000 & s_{2}^{2}=37,000 \\ \hline \end{array}$$ a. Do the data provide sufficient evidence to indicate that bulbs produced by the suspect line possess a larger variance in length of life than those produced by the line that is assumed to be in control? Use \(\alpha=.05\) b. Find the approximate observed significance level for the test and interpret its value.

A pharmaceutical manufacturer purchases a particular material from two different suppliers. The mean level of impurities in the raw material is approximately the same for both suppliers, but the manufacturer is concerned about the variability of the impurities from shipment to shipment. If the level of impurities tends to vary excessively for one source of supply, it could affect the quality of the pharmaceutical product. To compare the variation in percentage impurities for the two suppliers, the manufacturer selects ten shipments from each of the two suppliers and measures the percentage of impurities in the raw material for each shipment. The sample means and variances are shown in the accompanying table. $$\begin{array}{ll} \hline \text { Supplier A } & \text { Supplier B } \\ \hline \bar{y}_{1}=1.89 & \bar{y}_{2}=1.85 \\ s_{1}^{2}=.273 & s_{2}^{2}=.094 \\ n_{1}=10 & n_{2}=10 \\ \hline \end{array}$$ a. Do the data provide sufficient evidence to indicate a difference in the variability of the shipment impurity levels for the two suppliers? Test using \(\alpha=.10 .\) Based on the results of your test, what recommendation would you make to the pharmaceutical manufacturer? b. Find a \(90 \%\) confidence interval for \(\sigma_{\mathrm{B}}^{2}\) and interpret your results.

Why is the \(Z\) test usually inappropriate as a test procedure when the sample size is small?

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