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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. Explain why you chose the distribution you did for Exercise 10.24.

Short Answer

Expert verified
Use a two-sample t-test with \( \alpha = 0.05 \) to compare the means.

Step by step solution

01

State the Hypotheses

We want to test if the mean life spans of whites and nonwhites born in 1900 in the county are the same. Let \( \mu_W \) be the mean life span of whites and \( \mu_N \) be the mean life span of nonwhites. The null hypothesis \(H_0\) is that \( \mu_W = \mu_N \) and the alternative hypothesis \(H_a\) is that \( \mu_W eq \mu_N \).
02

Select the Significance Level and Test Distribution

We typically choose a significance level \( \alpha = 0.05 \). Because we are comparing means from two independent samples with unknown population standard deviations, we use the two-sample \( t \)-test.
03

Calculate the Test Statistic

The test statistic for two independent sample means is calculated using:\[ t = \frac{\bar{x}_W - \bar{x}_N - \Delta}{\sqrt{\frac{s_W^2}{n_W} + \frac{s_N^2}{n_N}}} \]where \( \bar{x}_W = 45.3 \), \( \bar{x}_N = 34.1 \), \( s_W = 12.7 \), \( s_N = 15.6 \), \( n_W = 124 \), \( n_N = 82 \), and \( \Delta = 0 \) under \( H_0 \). Substitute these values to calculate the \( t \)-statistic.
04

Determine the Degrees of Freedom

The degrees of freedom \( df \) for the two-sample \( t \)-test is approximated using the following formula:\[ df = \frac{\left(\frac{s_W^2}{n_W} + \frac{s_N^2}{n_N}\right)^2}{\frac{\left(\frac{s_W^2}{n_W}\right)^2}{n_W - 1} + \frac{\left(\frac{s_N^2}{n_N}\right)^2}{n_N - 1}} \]Calculate the \( df \) using the known values of \( s, n \).
05

Make a Decision

Using the \( t \)-statistic from Step 3 and the \( df \) from Step 4, find the critical \( t \)-value for \( \alpha = 0.05 \). If the \( t \)-statistic exceeds the critical \( t \)-value, we reject \( H_0 \); otherwise, we fail to reject \( H_0 \).
06

Conclusion

Based on the comparison in Step 5, conclude whether there is sufficient evidence to suggest that the mean life spans of whites and nonwhites in the county are different.

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Key Concepts

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

Two-Sample T-Test
When comparing the average life spans of two different groups, such as whites and nonwhites born in 1900, a two-sample t-test is an effective statistical method. This test is specifically designed to evaluate whether the means of two independent samples differ from each other significantly. In the context of our exercise, this involves checking if the life span of whites differs statistically from that of nonwhites. The two-sample t-test is appropriate in scenarios where we don't know the population standard deviations but can estimate them from our sample data.

To perform this test:
  • State the hypotheses: Our null hypothesis ( \( H_0 \)) asserts that the means are equal ( \( \mu_W = \mu_N \)), while the alternative hypothesis ( \( H_a \)) claims they are not ( \( \mu_W eq \mu_N \)).
  • Calculate the test statistic using the difference between sample means divided by the standard error, which takes into account both standard deviations and sample sizes.
This approach helps to assess whether any observed differences in means could be attributed to random sampling variability.
Significance Level
The significance level is a pivotal concept in hypothesis testing, most often denoted by the Greek letter \( \alpha \). It represents the probability threshold for rejecting the null hypothesis. In practical terms, it guards against false positives鈥攄rawing an incorrect conclusion that a difference exists when it actually does not. Typically, a significance level of 0.05 is utilized. This indicates a 5% risk of declaring a difference between means when there is none.

Choosing this level before conducting the test sets the groundwork for interpreting results. Should the test yield a \( p \)-value less than or equal to 0.05, it suggests that the observed data is inconsistent with the null hypothesis under the assumption it's true, providing enough evidence to consider rejecting it. Hence, the significance level not only influences decision-making but also reflects the tolerance for potential errors in the study.
Degrees of Freedom
Degrees of Freedom (df) in statistical tests indicate the number of independent values that can vary within a calculation, such as a statistical sample. In the context of the two-sample t-test, degrees of freedom are crucial for determining the critical \( t \)-value required to decide if the test statistic indicates a meaningful difference.

To calculate degrees of freedom for the two-sample t-test, we employ a somewhat complex formula accounting for two samples' sizes and variances. Understanding that this flexibility, represented by df, affects which critical \( t \)-value is applicable, emphasizes why it must be adequately computed. Accurate degrees of freedom ensure the reliability of conclusions drawn from the test, based on the test statistic and significance level.
Standard Deviation
Standard deviation is a fundamental statistical measure that quantifies the amount of variation or dispersion in a set of data values. In hypothesis testing, it plays a key role in calculating the test statistic by showing how data points differ from the mean.

For our exercise, the standard deviation helps express the inherent variability within the white and nonwhite life span samples. This quantity affects the standard error of the difference between means, influencing the size of the test statistic. Since larger standard deviations yield larger standard errors, they may decrease the likelihood of finding statistically significant differences between sample means, assuming the same sample size.

An understanding of standard deviation not only aids in interpreting the spread of data but is integral in determining the precision and reliability of statistical test outcomes.

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Most popular questions from this chapter

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. At a pre-conceived \(\alpha=0.05,\) what is your: a. Decision: b. Reason for the decision: c. Conclusion (write out in a complete sentence):

A recent year was randomly picked from 1985 to the present. In that year, there were 2,051 Hispanic students at Cabrillo College out of a total of 12,328 students. At Lake Tahoe College, there were 321 Hispanic students out of a total of 2,441 students. In general, do you think that the percent of Hispanic students at the two colleges is basically the same or different?

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. Which distribution (normal or Student's t) would you use for this hypothesis test?

Use the following information to answer the next five exercises. A study was conducted to test the effectiveness of a software patch in reducing system failures over a six-month period. Results for randomly selected installations are shown in Table 10.21. The 鈥渂efore鈥 value is matched to an 鈥渁fter鈥 value, and the differences are calculated. The differences have a normal distribution. Test at the 1% significance level. $$ \begin{array}{|l|l|l|l|l|l|l|}\hline \text { Installation } & {\mathbf{A}} & {\mathbf{B}} & {\mathbf{C}} & {\mathbf{D}} & {\mathbf{E}} & {\mathbf{F}} & {\mathbf{G}} & {\mathbf{H}} \\ \hline \text { Before } & {3} & {6} & {4} & {2} & {5} & {8} & {2} & {6} \\ \hline \text { After } & {1} & {5} & {2} & {0} & {1} & {0} & {2} & {2} \\ \hline\end{array} $$ State the null and alternative hypotheses.

Use the following information to answer the next five exercises. A doctor wants to know if a blood pressure medication is effective. Six subjects have their blood pressures recorded. After twelve weeks on the medication, the same six subjects have their blood pressure recorded again. For this test, only systolic pressure is of concern. Test at the 1% significance level. $$ \begin{array}{|l|l|l|l|l|l|}\hline \text { Patient } & {\mathbf{A}} & {\mathbf{B}} & {\mathbf{C}} & {\mathbf{D}} & {\mathbf{E}} & {\mathbf{F}} \\\ \hline \text { Before } & {161} & {162} & {165} & {162} & {166} & {171} \\\ \hline \text { After } & {158} & {159} & {166} & {160} & {167} & {169} \\\ \hline\end{array} $$ What is the sample mean difference?

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