Chapter 9: Problem 15
Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu=22 \\ H_{\mathrm{a}}: \mu \neq 22 \end{array} \\] A sample of 75 is used and the population standard deviation is \(10 .\) Compute the \(p\) -value and state your conclusion for each of the following sample results. Use \(\alpha=.01\) a. \(\bar{x}=23\) b. \(\bar{x}=25.1\) c. \(\quad \bar{x}=20\)
Short Answer
Step by step solution
Define the test statistic formula
Calculate the test statistic for \( \bar{x} = 23 \)
Find the p-value for \( \bar{x} = 23 \)
Conclusion for \( \bar{x} = 23 \)
Calculate the test statistic for \( \bar{x} = 25.1 \)
Find the p-value for \( \bar{x} = 25.1 \)
Conclusion for \( \bar{x} = 25.1 \)
Calculate the test statistic for \( \bar{x} = 20 \)
Find the p-value for \( \bar{x} = 20 \)
Conclusion for \( \bar{x} = 20 \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
z-test
- Given by the formula: \[ Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]
- Where \( \bar{x} \) is your sample mean, \( \mu \) is the known population mean, \( \sigma \) is the population standard deviation, and \( n \) is your sample size.
In the exercise you looked at, we calculated a z-score for different sample means. For instance, a sample mean of 23 resulted in a z-score of 0.866. This helps us understand how typical or unusual these average results are under the hypothesis.
p-value
- If your p-value is less than or equal to your significance level (often denoted by \( \alpha \)), you reject the null hypothesis.
- In the textbook example, we used \( \alpha = 0.01 \).
By calculating the p-value, researchers can decide whether their sample result is just a random fluke or it's really indicative of a different population mean. The p-value thus acts like a thermometer, telling you how 'hot' (or unusual) your sample data really is compared to the null expectation.
two-tailed test
- The two-tailed test considers deviations on both sides of the distribution tail.
- This means you must consider the extremities on both ends for possible outcomes.
With a two-tailed approach, you double the p-value of one tail test. For instance, if a z-score finds that the right tail probability is 0.1936, the two-tailed p-value becomes 0.3872 (as it incorporates both tails). Consider this like weighing all your options equally, whether they're above or below the assumed average.