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Chapter 8: Problem 12

Two populations have normal distributions. The first has population standard deviation 2 and the second has population standard deviation 3. A random sample of 16 measurements from the first population had a sample mean of \(20 .\) An independent random sample of 9 measurements from the second population had a sample mean of \(19 .\) Test the claim that the population mean of the first population exceeds that of the second. Use a \(5 \%\) level of significance. (a) What distribution does the sample test statistic follow? Explain. (b) State the hypotheses. (c) Compute \(\bar{x}_{1}-\bar{x}_{2}\) and the corresponding sample test statistic. (d) Find the \(P\) -value of the sample test statistic. (e) Conclude the test (f) The results.

Short Answer

Expert verified
The test statistic follows a Z-distribution. At 伪 = 0.05 , with a P-value of 0.1841 , we fail to reject H_0 ; insufficient evidence for the claim.

Step by step solution

01

Identify the Distribution

The sample test statistic follows a normal distribution because both population variances are known, and the Central Limit Theorem applies due to the sample sizes. Specifically, it follows a standard normal distribution (Z-distribution) because we are testing the difference between two means from normal populations with known variances.
02

State the Hypotheses

The null hypothesis ( H_0 ) states that the population mean of the first population ( ar{ ext{渭}}_1 ) does not exceed the population mean of the second population ( ar{ ext{渭}}_2 ), i.e., H_0: ar{ ext{渭}}_1 ext{ 鈮� } ar{ ext{渭}}_2 . The alternative hypothesis ( H_a ) states that the population mean of the first population is greater than that of the second, i.e., H_a: ar{ ext{渭}}_1 ext{ > } ar{ ext{渭}}_2 .
03

Calculate 9}ar{x}_1 - ar{x}_29}

Calculate the difference in sample means: ar{x}_1 - ar{x}_2 = 20 - 19 = 1 .
04

Compute Test Statistic

The test statistic is calculated using the formula:\[Z = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\]Substitute the given values:\[Z = \frac{(1) - 0}{\sqrt{\frac{2^2}{16} + \frac{3^2}{9}}} = \frac{1}{\sqrt{\frac{4}{16} + \frac{9}{9}}} = \frac{1}{\sqrt{0.25 + 1}} = \frac{1}{\sqrt{1.25}}\]This simplifies to:\[Z = \frac{1}{1.118} \approx 0.894\]
05

Find the P-value

Use the Z table to find the probability corresponding to Z = 0.894 . For the one-tailed test with Z = 0.894 , the P-value is 0.1855 . However, because we are testing if the mean of the first population is greater, we take 1 minus the table value to find the critical region P-value: P = 1 - 0.8159 = 0.1841 .
06

Draw Conclusion

Compare the P-value 0.1841 to the significance level 伪 = 0.05 . Since P > 伪 , we fail to reject the null hypothesis (H_0) . This indicates that there is insufficient evidence to support the claim that the population mean of the first population exceeds that of the second.
07

Result Summary

The test did not show a statistically significant difference in the population means at the 5% significance level. Therefore, based on the sample, we cannot conclude that the first population mean is greater than the second.

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Key Concepts

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

Normal Distribution
A normal distribution is a continuous probability distribution characterized by a symmetrical, bell-shaped curve. This means that most of the data points are concentrated around the mean, and there are fewer data points as you move away from the mean. When you hear about a normal distribution, think about the consistent patterns that many variables in nature and economics follow, such as height, test scores, and measurement errors.

Key features of a normal distribution include:
  • Symmetry around the mean.
  • The mean, median, and mode are all equal.
  • The area under the curve represents the total probability, which is equal to 1.
  • It is determined by two parameters: the mean () and the standard deviation ().
This exercise involves populations that follow a normal distribution, meaning the statistical methods used rely heavily on this property.
Z-distribution
The Z-distribution, also known as the standard normal distribution, is a specific type of normal distribution. It has a mean of 0 and a standard deviation of 1. This makes it a powerful tool for hypothesis testing, allowing statisticians to determine probabilities and critical values with ease by using standard normal tables.

In this exercise, we use the Z-distribution to compare two population means. When calculating the test statistic, a conversion to a Z-score allows us to go from the original data values to a standardized form. This is useful to easily determine whether the test statistic falls into the range of acceptance or rejection of the null hypothesis. This standardization makes the results applicable to other situations that share this distribution.
Population Mean
The population mean () represents the average of all the measurements in a population. It's a central measure of location and provides a summary statistic for the population as a whole. When performing hypothesis tests, particularly with normal distribution, the population mean serves as a baseline comparison point.

In our exercise, we are comparing the means of two populations to determine if one exceeds the other. In statistical hypothesis testing, the population mean is essential because it allows us to establish hypotheses and draw conclusions based on sample statistics relative to population characteristics. Understanding the population mean helps in making predictions and informed decisions based on your data.
Sample Mean
A sample mean () is an estimate of the population mean that is calculated from a subset of the entire population. It serves as a proxy when it's impractical or impossible to measure the whole population. By using the sample mean, researchers can make inferences about the population mean with a calculated level of confidence.

In hypothesis testing, the sample mean acts as a key statistic in assessing whether there is significant evidence to support a claim. In this task, the sample means are 20 and 19 for the first and second population samples, respectively. These values enable us to calculate the test statistic that provides insights into the population. By comparing sample means, the exercise demonstrates how statistical methods can help us infer results about larger populations from limited data.

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Most popular questions from this chapter

For one binomial experiment, 200 binomial trials produced 60 successes. For a second independent binomial experiment, 400 binomial trials produced 156 successes. At the \(5 \%\) level of significance, test the claim that the probability of success for the second binomial experiment is greater than that for the first. (a) Compute the pooled probability of success for the two experiments. (b) What distribution does the sample test statistic follow? Explain. (c) State the hypotheses. (d) Compute \(\hat{p}_{1}-\hat{p}_{2}\) and the corresponding sample test statistic. (e) Find the \(P\) -value of the sample test statistic. (f) Conclude the test. (g) The results.

The western United States has a number of four-lane interstate highways that cut through long tracts of wilderness. To prevent car accidents with wild animals, the highways are bordered on both sides with 12 -foot-high woven wire fences. Although the fences prevent accidents, they also disturb the winter migration pattern of many animals. To compensate for this disturbance, the highways have frequent wilderness underpasses designed for exclusive use by deer, elk, and other animals. In Colorado, there is a large group of deer that spend their summer months in a region on one side of a highway and survive the winter months in a lower region on the other side. To determine if the highway has disturbed deer migration to the winter feeding area, the following data were gathered on a random sample of 10 wilderness districts in the winter feeding area. Row \(B\) represents the average January deer count for a 5 -year period before the highway was built, and row \(A\) represents the average January deer count for a 5 -year period after the highway was built. The highway department claims that the January population has not changed. Test this claim against the claim that the January population has dropped. Use a \(5 \%\) level of significance. Units used in the table are hundreds of deer. \begin{tabular}{l|cccccccccc} \hline Wilderness District & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline\(B:\) Before highway & \(10.3\) & \(7.2\) & \(12.9\) & \(5.8\) & \(17.4\) & \(9.9\) & \(20.5\) & \(16.2\) & \(18.9\) & \(11.6\) \\ \hline\(A:\) After highway & \(9.1\) & \(8.4\) & \(10.0\) & \(4.1\) & \(4.0\) & \(7.1\) & \(15.2\) & \(8.3\) & \(12.2\) & \(7.3\) \\ \hline \end{tabular}

Suppose the \(P\) -value in a two-tailed test is \(0.0134\). Based on the same population, sample, and null hypothesis, and assuming the test statistic \(z\) is negative, what is the \(P\) -value for a corresponding left-tailed test?

Let \(x\) be a random variable that represents hemoglobin count (HC) in grams per 100 milliliters of whole blood. Then \(x\) has a distribution that is approximately normal, with population mean of about 14 for healthy adult women (see reference in Problem 17). Suppose that a female patient has taken 10 laboratory blood tests during the past year. The HC data sent to the patient's doctor are \(\begin{array}{llllllllll}15 & 18 & 16 & 19 & 14 & 12 & 14 & 17 & 15 & 11\end{array}\) i. Use a calculator with sample mean and sample standard deviation keys to verify that \(\bar{x}=15.1\) and \(s \approx 2.51\). ii. Does this information indicate that the population average \(\mathrm{HC}\) for this patient is higher than 14? Use \(\alpha=0.01\).

(a) For the same data and null hypothesis, is the \(P\) -value of a one-tailed test (right or left) larger or smaller than that of a two-tailed test? Explain. (b) For the same data, null hypothesis, and level of significance, is it possible that a one-tailed test results in the conclusion to reject \(H_{0}\) while a two-tailed test results in the conclusion to fail to reject \(H_{0}\) ? Explain. (c) For the same data, null hypothesis, and level of significance, if the conclusion is to reject \(H_{0}\) based on a two-tailed test, do you also reject \(H_{0}\) based on a one-tailed test? Explain. (d) If a report states that certain data were used to reject a given hypothesis, would it be a good idea to know what type of test (one-tailed or two-tailed) was used? Explain.
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