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Chapter 10: Problem 19

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. Is this a test of means or proportions?

Short Answer

Expert verified
This is a test of means.

Step by step solution

01

Understand the Type of Data

We are dealing with life spans measured in years for two groups: whites and nonwhites. This data is continuous, reflecting the age at death for individuals.
02

Identify the Nature of the Test

Given that we are examining the means (in particular, the mean life spans) of these two groups, we are conducting a test of means rather than a test of proportions. A test of proportions would involve categorical data, such as the percentage of a population with a particular characteristic.

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Key Concepts

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

Mean Comparison
Mean comparison is a crucial statistical method used to determine whether two groups have significantly different averages. In our scenario, we are comparing the mean life spans of whites and nonwhites born in 1900, based on a sample from a particular county. By examining whether these mean life spans are statistically different, we can glean insights about the variance in longevity across these groups.

Performing a mean comparison involves several key steps:
  • Determine the sample means: For whites, it is 45.3 years, and for nonwhites, it's 34.1 years.
  • Identify the population standard deviations and sample sizes: Here, the given standard deviations are 12.7 years for whites and 15.6 years for nonwhites.
  • Use a hypothesis test for the means: This typically involves setting up a null hypothesis assuming no difference in means, and an alternative hypothesis suggesting a difference.
Understanding the importance of a mean comparison helps in assessing differences in life expectancy or other metrics, thereby contributing to demographic studies and health analysis.
Standard Deviation
Standard deviation is a measure that helps quantify the amount of variation or dispersion within a data set. In our context, it tells us how much individual life spans differ from the mean life span in each group. A lower standard deviation indicates that the ages at death tend to be closer to the mean, while a higher standard deviation suggests wider variability.

In this exercise:
  • The standard deviation for whites is 12.7 years, suggesting a relatively moderate spread around the mean life expectancy of 45.3 years.
  • For nonwhites, the standard deviation is 15.6 years, which indicates more variability in life spans around the mean of 34.1 years.
Understanding standard deviation is essential for interpreting how consistent the life spans are within each racial group. It also plays a critical role in hypothesis testing and calculating the standard error, which is used to determine the reliability of the mean estimates.
Continuous Data Analysis
Continuous data includes any data set where values are not restricted to discrete categories but can exist anywhere on a continuous scale. Life span, in this example, is a type of continuous data as it can be measured in years with many possible values.

Key aspects of handling continuous data in hypothesis testing include:
  • Assessing normality: It's crucial to check if the data is normally distributed or if transformations are needed for analysis.
  • Using appropriate tests: For comparing means of two groups, we might use a t-test if the data meets the normality requirement and sample sizes are large enough.
  • Interpreting results: Calculating confidence intervals and p-values helps determine the statistical significance of any observed differences.
Analysis of continuous data provides detailed insights and allows for more nuanced interpretations compared to categorical data. This is particularly important in studies involving measurements like life expectancy, where understanding the exact figures is essential for effective analysis and conclusions.

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Most popular questions from this chapter

Use the following information to answer the next five exercises. Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. 15 pieces of each metal are being tested. Both populations have normal distributions. The following table is the result. It is believed that Alloy Zeta has a different melting point. $$\begin{array}{|l|l|l|}\hline & {\text { Sample Mean Melting Temperatures }\left(^{\circ} F\right)} & {\text { Population Standard Deviation }} \\\ \hline \text { Alloy Gamma } & {800}&{95} \\ \hline \text { Alloy zeta } & {900} &{105} \\ \hline\end{array}$$ At the 1% significance level, what is your conclusion?

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. $$ \begin{array}{|l|l|l|l|l|l|l|}\hline \text { Installation } & {\mathbf{A}} & {\mathbf{B}} & {\mathbf{C}} & {\mathbf{D}} & {\mathbf{E}} & {\mathbf{F}} & {\mathbf{G}} & {\mathbf{H}} \\ \hline \text { Before } & {3} & {6} & {4} & {2} & {5} & {8} & {2} & {6} \\ \hline \text { After } & {1} & {5} & {2} & {0} & {1} & {0} & {2} & {2} \\ \hline\end{array} $$ What is the random variable?

Use the following information to answer the next five exercises. Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. 15 pieces of each metal are being tested. Both populations have normal distributions. The following table is the result. It is believed that Alloy Zeta has a different melting point. $$\begin{array}{|l|l|l|}\hline & {\text { Sample Mean Melting Temperatures }\left(^{\circ} F\right)} & {\text { Population Standard Deviation }} \\\ \hline \text { Alloy Gamma } & {800}&{95} \\ \hline \text { Alloy zeta } & {900} &{105} \\ \hline\end{array}$$ Draw the graph of the p-value.

Use the following information to answer the next 15 exercises: Indicate if the hypothesis test is for a. independent group means, population standard deviations, and/or variances known b. independent group means, population standard deviations, and/or variances unknown c. matched or paired samples d. single mean e. two proportions f. single proportion It is thought that teenagers sleep more than adults on average. A study is done to verify this. A sample of 16 teenagers has a mean of 8.9 hours slept and a standard deviation of 1.2. A sample of 12 adults has a mean of 6.9 hours slept and a standard deviation of 0.6.

Use the following information to answer the next ten exercises. indicate which of the following choices best identifies the hypothesis test. a. independent group means, population standard deviations and/or variances known b. independent group means, population standard deviations and/or variances unknown c. matched or paired samples d. single mean e. two proportions f. single proportion A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. The population standard deviations are two pounds and three pounds, respectively. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet.
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