/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 In the following data pairs, \(A... [FREE SOLUTION] | 91Ó°ÊÓ

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In the following data pairs, \(A\) represents birth rate and \(B\) represents death rate per 1000 resident population. The data are paired by counties in the Midwest. A random sample of 16 counties gave the following information. (Reference: County and City Data Book, U.S. Department of Commerce.) \(\begin{array}{l|cccccccc} \hline \text { A: } & 12.7 & 13.4 & 12.8 & 12.1 & 11.6 & 11.1 & 14.2 & 15.1 \\\ \hline B: & 9.8 & 14.5 & 10.7 & 14.2 & 13.0 & 12.9 & 10.9 & 10.0 \\ \hline \\ \hline A: & 12.5 & 12.3 & 13.1 & 15.8 & 10.3 & 12.7 & 11.1 & 15.7 \\ \hline B: & 14.1 & 13.6 & 9.1 & 10.2 & 17.9 & 11.8 & 7.0 & 9.2 \\ \hline \end{array}\) Do the data indicate a difference (either way) between population average birth rate and death rate in this region? Use \(\alpha=0.01\).

Short Answer

Expert verified
Conduct a paired t-test; if calculated \( t \) > critical \( t \), reject \( H_0 \).

Step by step solution

01

Calculate the difference for each pair

For each county, calculate the difference between the birth rate (A) and the death rate (B). Let each difference be denoted as \( d_i = A_i - B_i \). Compute these differences for all 16 pairs.
02

Compute the mean and standard deviation of the differences

First, calculate the mean of the differences, \( \overline{d} \), by summing up all the differences and dividing by the number of counties, which is 16. Next, calculate the standard deviation \( s_d \) of the differences.
03

Formulate the null and alternative hypotheses

The null hypothesis \( H_0 \) is that there is no difference between the population average birth rate and death rate, i.e., the mean of the differences is zero, \( \mu_d = 0 \). The alternative hypothesis \( H_a \) is that there is a difference, i.e., \( \mu_d eq 0 \).
04

Calculate the test statistic

Use the formula for the test statistic: \( t = \frac{\overline{d}}{\frac{s_d}{\sqrt{n}}} \), where \( n \) is the number of data pairs, which is 16. Compute the value of \( t \) using the mean and standard deviation of the differences calculated earlier.
05

Determine the critical value and decision rule

For \( \alpha = 0.01 \) and 15 degrees of freedom (since \( n - 1 = 15 \)), find the critical \( t \) value from the t-distribution table for a two-tailed test. The decision rule is to reject the null hypothesis \( H_0 \) if the absolute value of the calculated \( t \) statistic is greater than the critical \( t \) value.
06

Make a conclusion

Compare the calculated \( t \) value to the critical \( t \) value. If \(|t|\) is greater, reject \( H_0 \), indicating there is a significant difference. Otherwise, do not reject \( H_0 \), indicating no significant difference.

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Key Concepts

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

Paired t-test
The paired t-test is a statistical method used to compare two related samples, often the same group twice under different conditions or at different times. In simple terms, it assesses whether the mean difference between paired observations is significantly different from zero.
  • This type of test is particularly useful when we are dealing with pairs of data, like in the situation of birth rates versus death rates across various counties.
  • It's important to note that the pairs are naturally matched, such as before-and-after measurements or, in this case, different yet related measurements from the same counties.

For our specific exercise with birth and death rates, we need to calculate the differences for each pair and then carry out the test to see if these differences, on average, differ significantly from zero. This helps determine if there is a real difference in rates across the population being studied.
Birth Rate vs Death Rate
Birth rate refers to the number of live births in a given year per 1,000 people in the population. On the other hand, the death rate indicates how many deaths occur per 1,000 people in the same timeframe. These rates are crucial in assessing population growth and health in a given area.
  • In our exercise, we are comparing these two metrics across counties to see if there is a statistically significant difference in their averages.
  • Understanding the balance between birth and death rates can offer insights into population dynamics, policy effectiveness, and future societal planning needs.

By analyzing the paired differences of these rates, we can explore the extent of any imbalance, guiding policy and intervention strategies.
t-distribution
The t-distribution is a type of probability distribution that is symmetric and bell-shaped, similar to the normal distribution but with heavier tails. This means it is more prone to producing values that fall far from its mean. It is particularly useful when dealing with small sample sizes.
  • In a paired t-test, the t-distribution helps us determine the critical value of the t-statistic for our hypothesis test.
  • This distribution becomes crucial when we have limited data, as it accounts for the uncertainty in the estimate of the standard deviation.

For our example, with a sample size of 16 counties, we use the t-distribution to find the critical value, helping us decide whether to reject or not reject the null hypothesis.
Two-tailed Test
A two-tailed test in statistical hypothesis testing checks for the possibility of an effect in two directions. In other words, it tests for differences that are either higher or lower than zero.
  • This is appropriate when we are unsure of what direction the difference might be, as is the case with birth vs. death rates.
  • Using a two-tailed test means we are assessing both the possibility that the birth rate could be significantly higher than the death rate, or vice versa.

In our exercise, using a two-tailed test involves considering any difference, not just one that is in a specific direction, which is key to thoroughly understanding these rates' comparative nature.

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Most popular questions from this chapter

To test \(\mu\) for an \(x\) distribution that is mound-shaped using sample size \(n \geq 30\), how do you decide whether to use the normal or Student's distribution?

Snow avalanches can be a real problem for travelers in the western United States and Canada. A very common type of avalanche is called the slab avalanche. These have been studied extensively by David McClung, a professor of civil engineering at the University of British Columbia. Slab avalanches studied in Canada had an average thickness of \(\mu=67 \mathrm{~cm}\) (Source: Avalancbe Handbook, by D. McClung and P. Schaerer). The ski patrol at Vail, Colorado, is studying slab avalanches in its region. A random sample of avalanches in spring gave the following thicknesses (in \(\mathrm{cm}\) ): \(\begin{array}{lllllll}59 & 51 & 76 & 38 & 65 & 54 & 49\end{array}\) 62 \(\begin{array}{llllllll}68 & 55 & 64 & 67 & 63 & 74 & 65 & 79\end{array}\) i. Use a calculator with mean and standard deviation keys to verify that \(\bar{x} \approx 61.8 \mathrm{~cm}\) and \(s \approx 10.6 \mathrm{~cm} .\) ii. Assume the slab thickness has an approximately normal distribution. Use \(1 \%\) level of significance to test the claim that the mean slab thickness in the Vail region is different from that in Canada.

Discuss each of the following topics in class or review the topics on your own. Then write a brief but complete essay in which you answer the following questions. (a) What is a null hypothesis \(H_{0} ?\) (b) What is an alternate hypothesis \(H_{1} ?\) (c) What is a type I error? a type II error? (d) What is the level of significance of a test? What is the probability of a type II error?

USA Today reported that about \(47 \%\) of the general consumer population in the United States is loyal to the automobile manufacturer of their choice. Suppose Chevrolet did a study of a random sample of 1006 Chevrolet owners and found that 490 said they would buy another Chevrolet. Does this indicate that the population proportion of consumers loyal to Chevrolet is more than \(47 \%\) ? Use \(\alpha=0.01\).

A study of fox rabies in southern Germany gave the following information about different regions and the occurrence of rabies in each region (Reference: B. Sayers, et al., "A Pattern Analysis Study of a Wildlife Rabies Epizootic," Medical Informatics 2:11-34). Based on information from this article, a random sample of \(n_{1}=16\) locations in region I gave the following information about the number of cases of fox rabies near that location. \(\begin{array}{llllllll}x_{1} \text { : Region I data } & 1 & 8 & 8 & 8 & 7 & 8 & 8 \\ & 3 & 3 & 3 & 2 & 5 & 1 & 4\end{array}\) A second random sample of \(n_{2}=15\) locations in region II gave the following information about the number of cases of fox rabies near that location. \(x_{2}\) : Region II data \(\quad 1\) i. Use a calculator with sample mean and sample standard deviation keys to verify that \(\bar{x}_{1}=4.75\) with \(s_{1} \approx 2.82\) in region \(\mathrm{I}\) and \(\bar{x}_{2} \approx 3.93\) with \(s_{2} \approx 2.43\) in region II. ii. Does this information indicate that there is a difference (either way) in the mean number of cases of fox rabies between the two regions? Use a \(5 \%\) level of significance. (Assume the distribution of rabies cases in both regions is mound-shaped and approximately normal.)

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