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Perform the following tests of hypotheses, assuming that the populations of paired differences are normally distributed. a. \(H_{0}: \mu_{d}=0, H_{1}: \mu_{d} \neq 0, n=9, \quad \bar{d}=6.7, \quad s_{d}=2.5, \quad \alpha=.10\) b. \(H_{0}: \mu_{d}=0, H_{1}: \mu_{d}>0, n=22, \bar{d}=14.8, s_{d}=6.4, \quad \alpha=.05\) c. \(H_{0}: \mu_{d}=0, H_{1}: \mu_{d}<0, n=17, \bar{d}=-9.3, s_{d}=4.8, \quad \alpha=.01\)

Short Answer

Expert verified
a. Do not reject the null hypothesis because the test statistic (8.04) is less than the critical value (±1.96). \n b. Reject the null hypothesis because the test statistic (8.66) is greater than the critical value (1.72). \n c. Reject the null Hypothesis because the test statistic (-7.27) is less than the critical value (-2.62).

Step by step solution

01

Compute the sample standard error

First, calculate the standard error for each hypothesis test. The standard error (SE) is given by \(\frac{s_d}{\sqrt{n}}\), where \(s_d\) is the sample standard deviation and \(n\) is the sample size.
02

Compute the test statistic

Next, calculate the test statistic for each hypothesis test. The test statistic is given by \(\frac{\bar{d} - \mu_d}{SE}\), where \(\bar{d}\) is the sample mean.
03

Determine the critical value(s)

Then, find the critical value or values. This will depend on whether you have a two-tailed test (\(\mu_d \neq 0\)) or a one-tailed test (\(\mu_d > 0\) or \(\mu_d < 0\)). Use a t-distribution table to find the critical value(s), using degrees of freedom (df) = \(n - 1\) and the given significance level \(\alpha\). If it is a two-tail test, divide the \(\alpha\) by 2 to find critical values in both tails. For a one-tail test, use the whole \(\alpha\) to find one critical value.
04

Compare the test statistic with the critical value and make a decision

Finally, compare the test statistic with the critical value. If the test statistic falls into the critical region (i.e., it is more extreme than the critical value), reject the null hypothesis (i.e., accept the alternative hypothesis). Otherwise, do not reject the null hypothesis.

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Key Concepts

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

Paired Differences
In hypothesis testing, examining paired differences involves comparing two related groups to determine any significant change or effect. This method is commonly used in before-and-after studies.
By analyzing the differences between paired observations, researchers can control for variability that might affect the results if examined separately. Paired difference tests are especially useful: - When the same subjects are measured under different conditions (e.g., performance before and after a training session). - In matching cases with controls. The main goal is to check if these paired differences arise purely by chance. If the population of these differences follows a normal distribution, then the subsequent steps using the t-distribution become valid.
T-Distribution
The t-distribution is a probability distribution that is used when the sample size is small, or when the population variance is unknown. It resembles the normal distribution but has thicker tails. These fat tails occur because there is a greater probability of extreme values when dealing with smaller samples. Key facts about the t-distribution: - It is symmetric and bell-shaped, like the normal distribution. - As the sample size increases, the t-distribution approaches the normal distribution. - It is defined by degrees of freedom, calculated as the sample size minus one (df = n - 1). In hypothesis testing involving paired differences, the t-distribution allows for the calculation of critical values, giving us a reference for deciding if a test statistic is extreme enough to reject the null hypothesis.
Test Statistic
The test statistic is a calculable value used to determine whether to reject the null hypothesis. For paired differences, the test statistic is based on the t-formula: \[ t = \frac{\bar{d} - \mu_d}{SE} \]Where:
  • \(\bar{d}\) is the sample mean of the differences.
  • \(\mu_d\) is the hypothesized mean difference, often set to zero.
  • \(SE\) is the standard error, \(\frac{s_d}{\sqrt{n}}\).
This statistic quantifies how far the actual sample mean is from the hypothesized population mean in standard error units.
A larger test statistic in absolute value indicates evidence against the null hypothesis.
Critical Value
The critical value is a threshold that the test statistic has to exceed for us to reject the null hypothesis. It is determined based on the significance level \(\alpha\), the degree of confidence (or confidence level), and the specific t-distribution.To find the critical value:
  • Use the t-distribution table or a statistical software.
  • Consider the degrees of freedom (df = n - 1).
  • Account for the type of test: one-tailed or two-tailed. For a two-tailed test, divide the significance level \(\alpha\) by 2.
When the test statistic is more extreme than this critical value (i.e., falls within the critical region), the evidence suggests that the null hypothesis should be rejected.

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Most popular questions from this chapter

The following information was obtained from two independent samples selected from two normally distributed populations with unknown but equal standard deviations. \(\begin{array}{llllllll}\text { Sample 1: } & 2.18 & 2.23 & 1.96 & 2.24 & 2.72 & 1.87 & 2.68 \\ & 2.15 & 2.49 & 2.05 & & & & \\ \text { Sample 2: } & 1.82 & 1.26 & 2.00 & 1.89 & 1.73 & 2.03 & 1.43 \\ & 2.05 & 1.54 & 2.50 & 1.99 & 2.13 & & \end{array}\) a. Let \(\mu_{1}\) be the mean of population 1 and \(\mu_{2}\) be the mean of population \(2 .\) What is the point estimate of \(\mu_{1}-\mu_{2} ?\) b. Construct a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Test at a \(2.5 \%\) significance level if \(\mu_{1}\) is lower than \(\mu_{2}\).

A sample of 1000 observations taken from the first population gave \(x_{1}=290\). Another sample of 1200 observations taken from the second population gave \(x_{2}=396\). a. Find the point estimate of \(p_{1}-p_{2}\). b. Make a \(98 \%\) confidence interval for \(p_{1}-p_{2}\). c. Show the rejection and nonrejection regions on the sampling distribution of \(\hat{p}_{1}-\hat{p}_{2}\) for \(H_{0}: p_{1}=p_{2}\) versus \(H_{1}: p_{1}

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According to a Bureau of Labor Statistics report released on March 25, 2015 , statisticians earn an average of \(\$ 84,010\) a year and accountants and auditors earn an average of \(\$ 73,670\) a year (www.bls. gov). Suppose that these estimates are based on random samples of 2000 statisticians and 1800 accountants and auditors. Further assume that the sample standard deviations of the annual earnings of these two groups are \(\$ 15,200\) and \(\$ 14,500\), respectively, and the population standard deviations are unknown and unequal for the two groups. a. Construct a \(98 \%\) confidence interval for the difference in the mean annual earnings of the two groups, statisticians and accountants and auditors. b. Using a \(1 \%\) significance level, can you conclude that the average annual earnings of statisticians is higher than that of accountants and auditors?

The manager of a factory has devised a detailed plan for evacuating the building as quickly as possible in the event of a fire or other emergency. An industrial psychologist believes that workers actually leave the factory faster at closing time without following any system. The company holds fire drills periodically in which a bell sounds and workers leave the building according to the system. The evacuation time for each drill is recorded. For comparison, the psychologist also records the evacuation time when the bell sounds for closing time each day. A random sample of 36 fire drills showed a mean evacuation time of \(5.1\) minutes with a standard deviation of \(1.1\) minutes. A random sample of 37 days at closing time showed a mean evacuation time of \(4.2\) minutes with a standard deviation of \(1.0\) minute. a. Construct a \(99 \%\) confidence interval for the difference between the two population means. b. Test at a \(5 \%\) significance level whether the mean evacuation time is smaller at closing time than during fire drills. Assume that the evacuation times at closing time and during fire drills have equal but unknown population standard deviations.

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