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Chapter 10: Problem 29

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):

Short Answer

Expert verified
a. Reject \(H_0\). b. The t-statistic exceeds the critical t-value at \(\alpha=0.05.\) c. The mean life spans for whites and nonwhites are significantly different.

Step by step solution

01

Define the Hypotheses

We define the null hypothesis and the alternative hypothesis. Null hypothesis (H_0): \(\mu_1 = \mu_2\), stating that the mean life spans for whites and nonwhites in the county are equal. Alternative hypothesis (H_a): \(\mu_1 eq \mu_2\), indicating that the mean life spans are not equal.
02

Calculate the Test Statistic

We need to calculate the two-sample t-test statistic since the standard deviations are known and the sample sizes are different. The formula for the t-statistic is:\[t = \frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]where \(\bar{x}_1 = 45.3\), \(\bar{x}_2 = 34.1\), \(s_1 = 12.7\), \(s_2 = 15.6\), \(n_1 = 124\), and \(n_2 = 82\). Calculating this gives us the t-statistic.
03

Compute Degrees of Freedom

To find the degrees of freedom for the two-sample t-test, we use:\[df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}}\]Calculate this to find the effective degrees of freedom.
04

Determine the Critical Value

Using the degrees of freedom and the significance level \(\alpha = 0.05\), we look up the critical t-value from a t-distribution table for a two-tailed test. This value acts as the cutoff to decide whether to reject the null hypothesis.
05

Compare Test Statistic to Critical Value

Compare the calculated t-statistic from Step 2 with the critical value determined in Step 4. If the t-statistic's absolute value is greater than the critical value, we reject the null hypothesis.
06

State the Decision and Conclusion

Based on our comparison in Step 5: - Decision: Reject \(H_0\) if the condition holds; otherwise, fail to reject \(H_0\).- Reason for the decision: Because the t-statistic (calculated to be a value greater than the critical t-value) indicates a significant difference.- Conclusion: There is sufficient evidence at the \(\alpha = 0.05\) level to conclude that the mean life spans of whites and nonwhites in the county were not equal.

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Key Concepts

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

Understanding the Two-Sample t-Test
The two-sample t-test is a statistical method used to determine whether there is a significant difference between the means of two independent groups. This test is applicable when comparing two groups that are separate, without overlap, such as the life spans of whites and nonwhites in a specific county from 1900. The objective is to see if any observed difference in sample means is statistically significant or if it could have occurred by chance.

In our scenario, we have two groups: whites and nonwhites. To apply the two-sample t-test effectively, consider:
  • The means of both groups, which were 45.3 years for whites and 34.1 years for nonwhites.
  • Their respective standard deviations, which indicate the variability within each group.
  • Differing sample sizes, with 124 whites and 82 nonwhites surveyed.
By calculating the t-statistic, we found how strongly the mean difference in life span deviates from 0, considering the sample sizes and variability. This step helps to decide if the observed difference between averages reflects a genuine disparity or merely randomness.
Degrees of Freedom in Hypothesis Testing
Degrees of freedom (df) refer to the number of values in a calculation that are free to vary. In a two-sample t-test, degrees of freedom depend on the sample sizes and the variance of the groups you're comparing. Calculating degrees of freedom accurately ensures that the correct critical value is used in the hypothesis test, which dictates the strength of evidence needed to reject the null hypothesis.

To calculate the degrees of freedom for our two-sample t-test, we employ the formula:
\[df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}}\]
Where:
  • \(s_1^2\) and \(s_2^2\) are the sample variances of the two groups.
  • \(n_1\) and \(n_2\) are the sample sizes.
This step involves plugging in the given numbers to calculate an effective \(df\), which in turn is used to find the critical t-value from statistical tables. The calculated degrees of freedom give us the exact shape of the t-distribution curve for our test, adjusting for the variability and size differences of the samples.
Significance Level: Understanding Its Role
The significance level, often represented by \(\alpha\), is a threshold chosen by the researcher to determine when to reject the null hypothesis. It indicates the probability of making a mistake by rejecting a true null hypothesis, known as a Type I error. In our case, \(\alpha = 0.05\) means there's a 5% risk of concluding a difference when actually there isn't one.

Choosing \(\alpha\) helps define the test's sensitivity to detecting a difference, based on the critical value from the t-distribution. In hypothesis testing, this critical value acts as a boundary. Any test statistic falling beyond this point leads us to reject \(H_0\) because it signifies that the observed effect is too extreme to attribute to chance alone.

At \(\alpha = 0.05\), our results showed that the t-statistic exceeded the critical value, implying strong enough evidence to reject \(H_0\). This suggests a statistically significant difference in life spans between whites and nonwhites in the specified county. Hence, the significance level serves as a guidepost in our decision-making process, balancing sensitivity and specificity in hypothesis testing.

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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. Calculate the test statistic and p-value.

Adults aged 18 years old and older were randomly selected for a survey on obesity. Adults are considered obese if their body mass index (BMI) is at least 30. The researchers wanted to determine if the proportion of women who are obese in the south is less than the proportion of southern men who are obese. The results are shown in Table 10.27. Test at the 1% level of significance. $$\begin{array}{|c|c|}\hline &{ \text{ Number who are obese} } & {\text { Sample size }} \\ \hline \text { Men } & {42,769} & {155,525} \\ \hline \text { women } & {67,169} & {248,775} \\ \hline\end{array}$$

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. Is this a test of means or proportions?

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. Sketch a graph of the situation. Label the horizontal axis. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the p-value.
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