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Chapter 10: Problem 30

Use the following information to answer the next 12 exercises: The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites. Does it appear that the means are the same? Why or why not?

Short Answer

Expert verified
The means are not the same; the hypothesis test indicates a significant difference between the groups.

Step by step solution

01

State the Hypotheses

We need to set up the null and alternative hypotheses. The null hypothesis (\[H_0\]) is that the mean life spans for whites and nonwhites are the same, which can be expressed as \[\mu_1 = \mu_2\]. The alternative hypothesis (\[H_a\]) is that the mean life spans are different, expressed as \[\mu_1 eq \mu_2\].
02

Determine Test Statistic

We use a two-sample t-test because we have two independent samples. The test statistic is calculated as \[t = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\], where \[\bar{x}_1 = 45.3\], \[\bar{x}_2 = 34.1\], \[s_1 = 12.7\], \[s_2 = 15.6\], \[n_1 = 124\], and \[n_2 = 82\]. Plug these values into the formula.
03

Calculate the Test Statistic

Let's compute the test statistic with the given values:\[ t = \frac{(45.3 - 34.1) - 0}{\sqrt{\frac{12.7^2}{124} + \frac{15.6^2}{82}}} \]\[ t = \frac{11.2}{\sqrt{\frac{161.29}{124} + \frac{243.36}{82}}} \]\[ t = \frac{11.2}{\sqrt{1.30 + 2.97}} \]\[ t = \frac{11.2}{1.93} \approx 5.80 \]
04

Determine the Degrees of Freedom

The degrees of freedom (\[df\]) for a two-sample t-test can be calculated using the formula for unequal variances:\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \]Substituting the values into the formula, compute the degrees of freedom.
05

Find the Critical Value and Make a Decision

Assume a significance level (\[\alpha\]) of 0.05. Find the critical value from the t-distribution table with the calculated degrees of freedom. If the absolute value of the test statistic is greater than the critical value, reject the null hypothesis. Otherwise, do not reject it.
06

Conclusion

Since the calculated test statistic \[t \approx 5.80\] is larger than the critical value from the t-distribution table at the \[\alpha = 0.05\] significance level, we reject the null hypothesis. This suggests that there is a significant difference in mean life spans between whites and nonwhites in this county.

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Key Concepts

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

Two-Sample T-Test
In statistics, when we want to compare the means of two independent groups, we often use a two-sample t-test. This test helps us determine if there is a significant difference between the means.
In our exercise, we are comparing the mean life spans of whites and nonwhites from the early 1900s. To use a two-sample t-test, the samples must be independent, and each group's data should follow a normal distribution.
  • The formula for the t-statistic is: \[ t = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]
  • \( \bar{x}_1 \) and \( \bar{x}_2 \) are the sample means, \( s_1 \) and \( s_2 \) are the sample standard deviations, \( n_1 \) and \( n_2 \) are the sample sizes for whites and nonwhites respectively.
In our case, this helped us calculate a t-statistic showing a significant difference. We then compare this to a critical value from the t-distribution.
Null Hypothesis
The null hypothesis, denoted as \( H_0 \), is a statement used in hypothesis testing that assumes no effect or no difference between groups. For every test, we begin by assuming that the null hypothesis is true. Our goal is to test whether the data provides enough evidence to reject this assumption.
In the context of our example, the null hypothesis states that the mean life spans of whites and nonwhites are equal. Mathematically, we express it as \( \mu_1 = \mu_2 \). All hypothesis tests aim to provide information on whether to reject the null hypothesis or fail to reject it based on the collected data and computation of statistics like the t-value.
Alternative Hypothesis
The alternative hypothesis, denoted \( H_a \), is what we consider when the null hypothesis is rejected. It proposes that there is a statistically significant effect or difference.
In our exercise, the alternative hypothesis suggests that the mean life spans of whites and nonwhites are not the same. This is written as \( \mu_1 eq \mu_2 \).
  • If our computed t-statistic exceeds the critical value from the t-distribution, we reject the null hypothesis in favor of the alternative.
  • This would imply a significant difference exists.
With a high calculated t-value from our data, we found evidence supporting the alternative hypothesis, confirming that the means significantly differ.
Degrees of Freedom
Degrees of freedom, often abbreviated as \( df \), represent the number of values in a statistical calculation that are free to vary.
In a two-sample t-test, degrees of freedom are essential to determine the critical value from the t-distribution table.
  • For our test with unequal sample variances, the formula is:\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \]
  • This accounts for the sample standard deviations and sizes of both groups involved.
Calculating the \( df \) allows identifying which critical t-value to use, leading to a conclusion about our hypotheses based on its comparison to the computed t-statistic.

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Most popular questions from this chapter

We are interested in whether children鈥檚 educational computer software costs less, on average, than children鈥檚 entertainment software. Thirty-six educational software titles were randomly picked from a catalog. The mean cost was \(31.14 with a standard deviation of \)4.69. Thirty-five entertainment software titles were randomly picked from the same catalog. The mean cost was \(33.86 with a standard deviation of \)10.87. Decide whether children鈥檚 educational software costs less, on average, than children鈥檚 entertainment software.

Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2. She takes random samples from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86 and 16.91, respectively. The mean and standard deviation for 37 statistics night students were 75.41 and 19.73. The 鈥渄ay鈥� subscript refers to the statistics day students. The 鈥渘ight鈥� subscript refers to the statistics night students. An appropriate alternative hypothesis for the hypothesis test is: a. \(\mu_{\text { day }}>\mu_{\text { night }}\) b. \(\mu_{\text { day }}<\mu_{\text { night }}\) c. \(\mu\) day \(=\mu_{\text { night }}\) d. \(\mu_{\text { day }} \neq \mu_{\text { night }}\)

Use the following information to answer the next five exercises. A researcher is testing the effects of plant food on plant growth. Nine plants have been given the plant food. Another nine plants have not been given the plant food. The heights of the plants are recorded after eight weeks. The populations have normal distributions. The following table is the result. The researcher thinks the food makes the plants grow taller. $$\begin{array}{|l|l|l|}\hline \text { Plant Group } & {\text { Sample Mean Height of Plants (inches) }} & {\text { Population Standard Deviation }} \\\ \hline \text { Food } & {16} & {2.5} \\ \hline \text { No food } & {14} & {1.5} \\ \hline\end{array}$$ Is the population standard deviation known or unknown?

Use the following information for the next five exercises. Two types of phone operating system are being tested to determine if there is a difference in the proportions of system failures (crashes). Fifteen out of a random sample of 150 phones with OS1 had system failures within the first eight hours of operation. Nine out of another random sample of 150 phones with OS2 had system failures within the first eight hours of operation. OS2 is believed to be more stable (have fewer crashes) than OS1. What is the random variable?

Use the following information for the next five exercises. Two types of phone operating system are being tested to determine if there is a difference in the proportions of system failures (crashes). Fifteen out of a random sample of 150 phones with OS1 had system failures within the first eight hours of operation. Nine out of another random sample of 150 phones with OS2 had system failures within the first eight hours of operation. OS2 is believed to be more stable (have fewer crashes) than OS1. What can you conclude about the two operating systems?
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