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Chapter 9: Problem 6

Decide whether you would reject or fail to reject the null hypothesis in the following situations: a. \(\overline{X_{D}}=3.50, s_{D}=1.10, n=12, \alpha=0.05\), two-tailed test b. \(95 \% \mathrm{CI}=(0.20,1.85)\) c. \(t=2.98, t *=-2.36\), one-tailed test to the left d. \(90 \% \mathrm{CI}=(-1.12,4.36)\)

Short Answer

Expert verified
a) Reject, b) Reject, c) Fail to reject, d) Fail to reject.

Step by step solution

01

Understand the Hypothesis Testing

Hypothesis testing is a method used to decide whether to reject or fail to reject a null hypothesis based on a sample statistic and confidence interval or critical value. The null hypothesis, often denoted as \(H_0\), is generally a statement of no effect or no difference, and we compare it with the alternative hypothesis (\(H_a\)).
02

Analyze Situation (a)

Given: \(\overline{X_{D}}=3.50, s_{D}=1.10, n=12, \alpha=0.05\) for a two-tailed test.- Calculate the standard error (SE): \(SE = \frac{s_D}{\sqrt{n}} = \frac{1.10}{\sqrt{12}} \approx 0.317\).- Find the critical t-value for \(\alpha = 0.05\) in a two-tailed test with 11 degrees of freedom using a t-table (approximately 2.201).- Compute the test statistic \(t\): \(t = \frac{\overline{X_{D}}-0}{SE} = \frac{3.50-0}{0.317} \approx 11.036\).- Since \(11.036 > 2.201\), reject the null hypothesis.
03

Analyze Situation (b)

Given: \(95\%\ \text{CI} = (0.20, 1.85)\).- The confidence interval does not contain 0. In a hypothesis test context, this implies that the null hypothesis \(H_0:\ \text{mean difference} = 0\) is not plausible.- Therefore, reject the null hypothesis.
04

Analyze Situation (c)

Given: \(t=2.98, t^*=-2.36\), for a one-tailed test to the left.- Expected critical value \(t^*\)=-2.36 is for rejecting when observing extreme values on the left.- Since provided \(t=2.98\) is on the opposite side of \(t^*=-2.36\), fail to reject the null hypothesis.
05

Analyze Situation (d)

Given: \(90\%\ \text{CI} = (-1.12, 4.36)\).- The confidence interval contains 0, suggesting the mean difference could be 0, and thus there is insufficient evidence to contradict the null hypothesis.- Therefore, fail to reject the null hypothesis.

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Key Concepts

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

Null Hypothesis
The null hypothesis, often symbolized as \(H_0\), is a foundational concept in hypothesis testing. It represents a default position that there is no effect or no difference in the population. For example, when examining the effectiveness of a new medication, \(H_0\) might assert that the medication has no effect compared to a placebo. When conducting a hypothesis test, researchers seek to determine whether there is enough evidence to reject this hypothesis in favor of an alternative hypothesis, which proposes some kind of effect or difference.
  • The null hypothesis serves as a starting point for statistical testing, allowing researchers to apply mathematical techniques to evaluate the plausibility of a hypothesis.
  • Rejecting \(H_0\) usually suggests that the alternative hypothesis \(H_a\) might be true, indicating a meaningful effect or difference exists.
  • Failing to reject \(H_0\) means there is not enough evidence to support the alternative hypothesis, but it does not necessarily prove \(H_0\) is true.
By setting up these hypotheses, researchers can use statistical evidence to make informed conclusions about their studies.
Confidence Interval
A confidence interval is a range of values, derived from sample data, that is likely to contain the true value of a population parameter. It offers an estimate of uncertainty around a measurement and is often expressed at a confidence level, such as 95% or 90%. This level indicates the probability that the interval will capture the true parameter value if the same population is sampled multiple times. Confidence intervals play a crucial role in hypothesis testing:
  • If a confidence interval for a mean difference does not contain 0, it suggests that a true difference likely exists, leading researchers to reject the null hypothesis.
  • If the interval includes 0, it implies that there is no significant difference, thus failing to reject the null hypothesis.
  • They provide a visual and numerical summary of the uncertainty in an estimate, which can be more informative than simply relying on p-values.
For example, in the provided exercise, a 95% CI of (0.20, 1.85) does not contain 0, leading to the rejection of the null hypothesis. This suggests a significant difference was found.
Test Statistic
The test statistic is a standardized value calculated from sample data during a hypothesis test. It is used to determine whether to reject the null hypothesis. Common test statistics include t-statistic, z-statistic, chi-square, and F-statistic. These statistics help compare observed data to what is expected under the null hypothesis.Calculating the test statistic involves the following steps:
  • First, calculate the standard error, which measures how much sample data is expected to fluctuate around the population parameter.
  • Next, determine the test statistic by comparing the observed sample mean or proportion to the hypothesized population value using the formula: \( t = \frac{\overline{X} - \mu}{SE} \), where \(\overline{X}\) is the sample mean, \(\mu\) is the population mean, and \(SE\) is the standard error.
  • Compare the test statistic to critical values from relevant statistical tables (such as the t-table or z-table) to decide whether to reject \(H_0\).
In context, when a test statistic exceeds the critical value in absolute terms, the null hypothesis can be rejected. In the exercise, the computed t-statistic was 11.036, which was significantly higher than the critical value 2.201, resulting in the rejection of \(H_0\).

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Most popular questions from this chapter

Name 3 research questions that could be addressed using a dependent samples \(t\) -test.

Why is the null hypothesis for a dependent-samples \(t\) -test always \(\mu_{D}=0 ?\)

What is the difference between a 1-sample \(t\) -test and a dependent-samples \(t\) -test? How are they alike?

Construct confidence intervals from a mean of \(\overline{X_{D}}=1.25\), standard error of \(s_{\overline{X_{D}}}=0.45\), and \(d f=10\) at the \(90 \%, 95 \%\), and \(99 \%\) confidence level. Describe what happens as confidence changes and whether to reject \(H_{0}\).

You want to know if an employee's opinion about an organization is the same as the opinion of that employee's boss. You collect data from 18 employee- supervisor pairs and code the difference scores so that positive scores indicate that the employee has a higher opinion and negative scores indicate that the boss has a higher opinion (meaning that difference scores of 0 indicate no difference and complete agreement). You find that the mean difference score is \(\overline{X_{D}}=-3.15\) with a standard deviation of \(s_{D}=1.97\). Test this hypothesis at the \(\alpha=0.01\) level. Construct confidence intervals from a mean of \(\overline{X_{D}}=1.25\), standard error of \(s_{\overline{X_{D}}}=0.45\), and \(d f=10\) at the \(90 \%, 95 \%\), and \(99 \%\) confidence level.
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