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Conduct the following tests of hypotheses, assuming that the populations of paired differences are normally distributed a. \(H_{0}: \mu_{d}=0, \quad H_{1}: \mu_{d} \neq 0, \quad n=26, \quad \bar{d}=9.6, \quad s_{d}=3.9, \quad \alpha=.05\) b. \(H_{0}: \mu_{d}=0, \quad H_{1}: \mu_{d}>0, \quad n=15, \quad \bar{d}=8.8, \quad s_{d}=4.7, \quad \alpha=.01\) c. \(H_{0}: \mu_{d}=0, \quad H_{1}: \mu_{d}<0, \quad n=20, \quad \bar{d}=-7.4, \quad s_{d}=2.3, \quad \alpha=.10\)

Short Answer

Expert verified
For part (a), (b), and (c), the comparison of the calculated t-value and the critical value from the t-distribution table will determine whether to reject or not reject the null hypothesis.

Step by step solution

01

Set up Hypotheses

The null hypothesis \(H_0\) is that the mean of the paired differences is zero. The alternative hypothesis \(H_1\) differs for each part of the exercise: (a) \(H_1: \mu_{d} \neq 0\) (a two-tailed test), (b) \(H_1: \mu_{d}>0\) (a one-tailed test with the significant area in the right tail), (c) \(H_1: \mu_{d}<0\) (a one-tailed test with the significant area in the left tail).
02

Calculate Test Statistic

Calculate test statistic by using formula for the t value: \(t = \frac{\bar{d} - \mu_{d}}{\frac{s_{d}}{\sqrt{n}}}\). Calculate this for each part.
03

Determine Critical Value

Determine the critical value using the t-distribution table. The number of degrees of freedom is \(n-1\). The critical value corresponds to the provided significance level \(\alpha\). The critical value differs for each part depending on whether the test is one-tailed or two-tailed.
04

Make Decision

Compare the test statistic with the critical value. If the test statistic is in the significance area (for one-tailed tests) or outside of the interval between the negative and positive critical value (for two-tailed tests), reject the null hypothesis. Else do not reject the null hypothesis. Do this for each part.

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Key Concepts

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

Paired Samples t-Test
When we have paired samples, we typically want to compare two related groups to see if there's a significant difference between them. The paired samples t-test is the ideal method for this analysis. Imagine you have a group of students taking a test before and after a preparation course. You want to know if the course made any improvement in scores. That's where the paired samples t-test comes into play.

Before running the test, ensure that your data are normally distributed. What makes this test special is that it looks at the differences within each pair rather than comparing individual group scores. This helps eliminate variability that might come from external factors.

  • The sample sizes in our examples are small, so using the t-distribution is more appropriate than using the normal distribution.
  • This test can be one-tailed or two-tailed, depending on whether you're looking for differences in a specific direction.
When analyzing data through a paired samples t-test, we're trying to establish whether the mean difference between the paired observations is statistically significant.
Null Hypothesis
The null hypothesis, denoted as \(H_0\), is a fundamental concept in hypothesis testing. It posits that there is no effect or no difference in the context that you're testing. In every hypothesis test, such as the ones mentioned in our original exercise, we start by assuming the null hypothesis is true. This is because it provides a baseline to measure any observed effects against.

For the paired samples scenarios we've discussed, the null hypothesis is that the mean difference between the paired observations is zero, symbolically expressed as \(H_0: \mu_d = 0\). This implies that any change or difference observed is due to randomness or sampling variation.

The purpose of hypothesis testing is to determine whether we have enough evidence to reject this null hypothesis in favor of an alternative hypothesis \(H_1\), which suggests an actual effect or difference exists. Only if the data show a strong enough departure from what we expect under \(H_0\), we consider the alternative.
Test Statistic
The test statistic is a calculated value used to compare the observed data to what is expected under the null hypothesis. In the context of our paired samples t-test, the test statistic allows us to measure how many standard deviations our sample mean difference \(\bar{d}\) is away from the mean difference of zero under the null hypothesis.

The formula to calculate the t-statistic is:

\[ t = \frac{\bar{d} - 0}{\frac{s_d}{\sqrt{n}}} \]

  • \(\bar{d}\) is the mean of the differences.
  • \(s_d\) is the standard deviation of the differences.
  • \(n\) is the number of pairs.
In our exercise, the t-statistic is computed for each part separately, and it gives us a standardized way of figuring out how far the mean difference is from zero, under the assumptions of the null hypothesis.
Critical Value
The critical value is a threshold that helps us decide whether the test statistic provides enough evidence against the null hypothesis. It is obtained from a table corresponding to the t-distribution, which considers the confidence level we set (usually derived from the significance level \(\alpha\)).

We determine the critical value by taking into account:
  • The number of degrees of freedom, which is calculated as \(n-1\) for paired samples.
  • The level of significance \(\alpha\), which represents the probability of rejecting the null hypothesis when it is actually true. For example, \(\alpha = 0.05\) or \(0.01\).
  • Whether the test is one-tailed or two-tailed.
If the test statistic falls into a critical region defined by the critical value, we reject the null hypothesis. For example, in part (a) with \(\alpha=0.05\) for a two-tailed test, we look at both ends of the t-distribution. Understanding how to interpret these values is key to making the correct statistical decisions.

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Most popular questions from this chapter

A random sample of nine students was selected to test for the effectiveness of a special course designed to improve memory. The following table gives the scores in a memory test given to these students before and after this course. $$ \begin{array}{l|lllllllll} \hline \text { Before } & 43 & 57 & 48 & 65 & 81 & 49 & 38 & 69 & 58 \\ \hline \text { After } & 49 & 56 & 55 & 77 & 89 & 57 & 36 & 64 & 69 \\ \hline \end{array} $$ a. Construct a \(95 \%\) confidence interval for the mean \(\mu_{d}\) of the population paired differences, where a paired difference is defined as the difference between the memory test scores of a student before and after attending this course. b. Test at the \(1 \%\) significance level whether this course makes any statistically significant improvement in the memory of all students.

A sample of 500 observations taken from the first population gave \(x_{1}=305\). Another sample of 600 observations taken from the second population gave \(x_{2}=348\). a. Find the point estimate of \(p_{1}-p_{2}\). b. Make a \(97 \%\) 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 \(H_{1}: p_{1}>p_{2} .\) Use a significance level of \(2.5 \% .\) d. Find the value of the test statistic \(z\) for the test of part \(\mathrm{c}\). e. Will you reject the null hypothesis mentioned in part \(\mathrm{c}\) at a significance level of \(2.5 \%\) ?

Find the following confidence intervals for \(\mu_{d}\), assuming that the populations of paired differences are normally distributed. a. \(n=11, \quad d=25.4, \quad s_{d}=13.5\), confidence level \(=99 \%\) b. \(n=23, \quad \bar{d}=13.2, \quad s_{d}=4.8, \quad\) confidence level \(=95 \%\) c. \(n=18, \quad \bar{d}=34.6, \quad s_{d}=11.7, \quad\) confidence level \(=90 \%\)

A state that requires periodic emission tests of cars operates two emissions test stations, \(\mathrm{A}\) and \(\mathrm{B}\), in one of its towns. Car owners have complained of lack of uniformity of procedures at the two stations, resulting in different failure rates. A sample of 400 cars at Station A showed that 53 of those failed the test; a sample of 470 cars at Station B found that 51 of those failed the test. a. What is the point estimate of the difference between the two population proportions? b. Construct a \(95 \%\) confidence interval for the difference between the two population proportions. c. Testing at the \(5 \%\) significance level, can you conclude that the two population proportions are different? Use both the critical-value and the \(p\) -value approaches.

A car magazine is comparing the total repair costs incurred during the first three years on two sports cars, the T-999 and the XPY. Random samples of 45 T-999s and 51 XPYs are taken. All 96 cars are 3 years old and have similar mileages. The mean of repair costs for the 45 T-999 cars is \(\$ 3300\) for the first 3 years. For the 51 XPY cars, this mean is \(\$ 3850 .\) Assume that the standard deviations for the two populations are \(\$ 800\) and \(\$ 1000\), respectively. a. Construct a \(99 \%\) confidence interval for the difference between the two population means. b. Using the \(1 \%\) significance level, can you conclude that such mean repair costs are different for these two types of cars? c. What would your decision be in part \(\mathrm{b}\) if the probability of making a Type I error were zero? Explain.

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