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Two populations have mound-shaped, symmetric distributions. A random sample of 16 measurements from the first population had a sample mean of 20 , with sample standard deviation \(2 .\) An independent random sample of 9 measurements from the second population had a sample mean of 19 , with sample standard deviation \(3 .\) 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) Estimate the \(P\) -value of the sample test statistic. (e) Conclude the test. (f) The results.

Short Answer

Expert verified
We use a t-distribution; do not reject the null hypothesis.

Step by step solution

01

Identify the Distribution

Since both populations have symmetric mound-shaped distributions (i.e., they are normally distributed or have similar characteristics), we use a **t-distribution** to analyze the difference between the two independent sample means. The degrees of freedom will be calculated using the smaller of the two sample sizes minus one.
02

State the Hypotheses

The null hypothesis (H_0) states that there is no difference in the population means, \(\mu_1 \leq \mu_2 \). The alternative hypothesis (H_a) is that the population mean of the first exceeds the second, \(\mu_1 > \mu_2 \).
03

Compute Sample Mean Difference and Test Statistic

First, find the difference in sample means: \(\bar{x}_1 - \bar{x}_2 = 20 - 19 = 1 \). Then, calculate the standard error of the difference in means using the formula \(SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{2^2}{16} + \frac{3^2}{9}} = \sqrt{0.25 + 1} = \sqrt{1.25} \approx 1.118 \). The test statistic \(t\) is given by \(t = \frac{(\bar{x}_1 - \bar{x}_2)}{SE} = \frac{1}{1.118} \approx 0.894 \).
04

Determine the P-value

For a right-tailed test with \(u = 8\) (assuming the degrees of freedom from the smallest sample \(n_2 - 1 = 8\)), use a t-table or software to find the \(P\)-value associated with \(t = 0.894 \). Approximate \(P\)-value is greater than 0.05.
05

Conclude the Test

Since the \(P\)-value is greater than 0.05, we do not reject the null hypothesis. This implies there is insufficient statistical evidence to support the claim that the population mean of the first population exceeds that of the second.
06

Summary of Results

The results of the test indicate that there is no significant evidence at the 5% significance level to state that the mean of the first population is greater than that of the second. Therefore, we do not have enough evidence to support the claim.

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Key Concepts

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

t-distribution
The **t-distribution** is a probability distribution used in statistics when we have small sample sizes and the population standard deviation is unknown. It's particularly helpful in hypothesis testing when dealing with sample means. Unlike the normal distribution, the t-distribution has heavier tails, which makes it more adaptable to small sample situations.

This distribution is symmetric and bell-shaped, similar to the normal distribution, but it's broader. The exact shape of the t-distribution depends on the degrees of freedom, which usually equals the sample size minus one. In our exercise, since we have a smaller sample from the second population with 9 observations, the degrees of freedom would be 9 - 1 = 8. This is important because as the degrees of freedom increase, the t-distribution closely resembles a normal distribution. In hypothesis testing for this exercise, we use the t-distribution to find critical values needed for testing our hypotheses.
sample mean
The **sample mean** is the average value of a sample, and it's a central point in statistics. It's the sum of all observations divided by the number of observations. In our exercise, the sample mean for the first population is 20, and for the second population, it's 19.

Sample means provide an estimation of the population mean and are crucial in comparing two or more groups. When we find the difference between the sample means—in this case, 20 - 19 = 1—we are considering the observed effect size. This difference helps us form our test statistic and see if there's any significant difference between the two population means.

Determining whether this observed difference is due to random chance or reflects a true difference in population means is what hypothesis testing evaluates.
p-value
The **p-value** is a statistical figure that tells us the probability that the observed data occurred by chance, assuming the null hypothesis is true. In simple terms, it's the evidence against a null hypothesis.

In this context of hypothesis testing, a smaller p-value indicates stronger evidence against the null hypothesis. We usually compare the p-value to a predefined significance level, often 0.05, to make a decision. In our exercise, after calculating the test statistic, we find the p-value is greater than 0.05. This implies we don't have enough evidence to reject the null hypothesis at the 5% significance level.

Thus, the p-value plays a critical role in decision-making during hypothesis testing, guiding whether we accept or reject the null hypothesis.
standard error
The **standard error** measures how much the sample mean deviates from the actual population mean. It's an essential element in calculating the test statistic for hypothesis testing and gives a sense of the uncertainty in our sample statistic.

We calculate the standard error when working with differences between two independent sample means using the formula: \[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]Here, - \( s_1 \) and \( s_2 \) are the sample standard deviations, and- \( n_1 \) and \( n_2 \) are the sample sizes.

For this exercise, the standard error turned out to be approximately 1.118. A smaller standard error indicates that the sample mean is a more precise estimate of the population mean. Knowing this helps us grasp the reliability of our findings and the variability of the sample means.

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Most popular questions from this chapter

Consider a set of data pairs. What is the first step in processing the data for a paired differences test? What is the formula for the sample test statistic \(t ?\) Describe each symbol used in the formula.

When using a Student's \(t\) distribution for a paired differences test with \(n\) data pairs, what value do you use for the degrees of freedom?

A Michigan study concerning preference for outdoor activities used a questionnaire with a 6 -point Likert-type response in which 1 designated "not important" and 6 designated "extremely important." A random sample of \(n_{1}=46\) adults were asked about fishing as an outdoor activity. The mean response was \(\bar{x}_{1}=4.9 .\) Another random sample of \(n_{2}=51\) adults were asked about camping as an outdoor activity. For this group, the mean response was \(\bar{x}_{2}=4.3 .\) From previous studies, it is known that \(\sigma_{1}=1.5\) and \(\sigma_{2}=1.2 .\) Does this indicate a difference (either way) regarding preference for camping versus preference for fishing as an outdoor activity? Use a \(5 \%\) level of significance. Note: \(\mathrm{A}\) Likert scale usually has to do with approval of or agreement with a statement in a questionnaire. For example, respondents are asked to indicate whether they "strongly agree," "agree," "disagree," or "strongly disagree" with the statement.

The body weight of a healthy 3 -month-old colt should be about \(\mu=60 \mathrm{~kg}\) (Source: The Merck Veterinary Mamual, a standard reference manual used in most veterinary colleges). (a) If you want to set up a statistical test to challenge the claim that \(\mu=60 \mathrm{~kg}\), what would you use for the null hypothesis \(H_{0} ?\) (b) In Nevada, there are many herds of wild horses. Suppose you want to test the claim that the average weight of a wild Nevada colt ( 3 months old) is less than \(60 \mathrm{~kg}\). What would you use for the alternate hypothesis \(H_{1} ?\) (c) Suppose you want to test the claim that the average weight of such a wild colt is greater than \(60 \mathrm{~kg}\). What would you use for the alternate hypothesis? (d) Suppose you want to test the claim that the average weight of such a wild colt is different from \(60 \mathrm{~kg}\). What would you use for the alternate hypothesis? (e) For each of the tests in parts (b), (c), and (d), would the area corresponding to the \(P\) -value be on the left, on the right, or on both sides of the mean? Explain your answer in each case.

What terminology do we use for the probability of rejecting the null hypothesis when it is, in fact, false?

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