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Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu=15 \\ H_{\mathrm{a}}: \mu \neq 15 \end{array} \\] A sample of 50 provided a sample mean of \(14.15 .\) The population standard deviation is 3 a. Compute the value of the test statistic. b. What is the \(p\) -value? c. \(\quad\) At \(\alpha=.05,\) what is your conclusion? d. What is the rejection rule using the critical value? What is your conclusion?

Short Answer

Expert verified
Reject \(H_0\); \(\mu \neq 15\).

Step by step solution

01

State the Hypotheses

The null hypothesis \(H_0\) states that \(\mu = 15\) while the alternative hypothesis \(H_a\) states that \(\mu eq 15\). This is a two-tailed test.
02

Calculate the Test Statistic

The formula to calculate the test statistic \(z\) for a sample mean is given by: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]Substitute the given values: \(\bar{x} = 14.15\), \(\mu = 15\), \(\sigma = 3\), and \(n = 50\):\[ z = \frac{14.15 - 15}{\frac{3}{\sqrt{50}}} = \frac{-0.85}{\frac{3}{7.071}} = \frac{-0.85}{0.424} \approx -2.0047 \]
03

Find the p-value

For a two-tailed test, the \(p\)-value is calculated based on the test statistic and the standard normal distribution. Using a standard normal distribution table or calculator:\[ p = 2 \times P(Z > |z|) \]Since \(z = -2.0047\),\[ p \approx 2 \times P(Z > 2.0047) \approx 2 \times 0.0227 = 0.0454 \]
04

Compare p-value with Significance Level

The significance level \(\alpha\) is 0.05. Compare the \(p\)-value to \(\alpha\): Since \(0.0454 < 0.05\), the \(p\)-value is less than \(\alpha\).
05

Conclusion for p-value Method

Since the \(p\)-value is less than \(\alpha\), we reject the null hypothesis \(H_0\). There is sufficient evidence to suggest that \(\mu eq 15\).
06

Determine the Critical Value

For \(\alpha = 0.05\) in a two-tailed test, the critical values are \(z = \pm 1.96\).
07

Compare Test Statistic with Critical Value

The test statistic \(z = -2.0047\) is less than \(-1.96\), falls in the critical region.
08

Conclusion for Critical Value Method

Since the test statistic \(-2.0047\) falls in the critical region, we reject the null hypothesis \(H_0\). There is enough evidence to suggest \(\mu eq 15\).

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis, often represented as \(H_0\), is a statement that assumes no effect or no difference in a specific situation. It serves as a baseline or benchmark for statistical testing.
For example, when the null hypothesis states \(\mu = 15\), it suggests that the mean of the population from which the sample is drawn is 15.
The null hypothesis is tested against an alternative hypothesis, symbolized by \(H_a\). The alternative hypothesis claims that there is a significant effect or difference, implying that the mean is not equal to 15 in the given exercise.
Understanding the null hypothesis is crucial because the conclusions of hypothesis tests revolve around whether there is sufficient evidence to reject this hypothesized statement or not.
p-value
The p-value in hypothesis testing is a critical concept for understanding the strength of the evidence against the null hypothesis. It represents the probability of obtaining test results at least as extreme as the ones observed, under the assumption that the null hypothesis is true.
In our example, a p-value of approximately 0.0454 was calculated, which tells us that there is a 4.54% chance of observing a sample mean as extreme as 14.15, assuming the population mean is actually 15.
The lower the p-value, the stronger the evidence against the null hypothesis. In hypothesis testing, we often compare the p-value to a pre-determined significance level, \(\alpha\), such as 0.05.
If the p-value is less than \(\alpha\), we reject the null hypothesis, suggesting strong evidence that the null hypothesis may not be true.
Test Statistic
The test statistic is a standardized value that helps determine how far the sample statistic is from the population parameter, under the null hypothesis. In hypothesis testing, computing the test statistic is a crucial step.
For the mean, the most common formula used is:
  • \( z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \)
Where:
  • \(\bar{x}\) is the sample mean
  • \(\mu\) is the population mean
  • \(\sigma\) is the population standard deviation
  • \(n\) is the sample size
In the exercise, substituting the given values results in a test statistic \(z = -2.0047\).
This value indicates how many standard deviations the sample mean is away from the assumed population mean. A further distance implies stronger evidence against the null hypothesis.
Critical Value
Critical values help define the boundaries of the rejection region in hypothesis tests. They are predetermined cutoff points decided by the significance level \(\alpha\), dictating the extreme areas of the distribution where the null hypothesis will be rejected.
In a two-tailed test with \(\alpha = 0.05\), the critical values are determined to be \(z = \pm 1.96\).
These values set the boundaries of "usual" vs. "unusual" results under the null hypothesis, sending any test statistic that falls beyond them into the rejection region.
In this case, the calculated test statistic \(z = -2.0047\) fell within this critical region, indicating that the sample evidence was strong enough to reject the null hypothesis.

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Most popular questions from this chapter

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