Please See The Attached And Follow Instructions Post Two Of
Please See The Attached And Follow Instructionspost Two Of The Six Exe
Please see the attached and follow instructions. Post two of the six exercises from the list for Hypothesis Testing from Chapter 9 of Homework section from Illowsky. Post both exercises as an attachment in a single thread for grading. Include the chapter number and exercise number in the title of the posting, e.g., 9.5.75 & 9.5.115. Use a level of 5% unless otherwise stated. If the sample size is less than n=30, use a "t" distribution instead of "Z." Do not compute p-value for "t" distribution problems; instead, compare the test value of "t" with the critical value corresponding to the given level.
Paper For Above instruction
Introduction
Hypothesis testing is a fundamental aspect of inferential statistics, allowing researchers to make decisions about population parameters based on sample data. The exercises from Chapter 9 of Illowsky’s homework guide focus on applying hypothesis testing procedures, particularly using t-tests and z-tests under different sample sizes and significance levels. This paper addresses two exercises from the specified chapter, providing detailed processes of formulating hypotheses, calculating test statistics, and comparing these to critical values to reach conclusions.
Exercise 1: Hypothesis Testing with Small Sample Size (Using t-distribution)
Suppose Exercise 9.5.74 involves testing a claim about a population mean with a sample size less than 30. To illustrate, consider a scenario where the sample mean is 50, the hypothesized population mean is 45, the sample standard deviation is 8, and the sample size is 25. The null hypothesis (H₀) states that the population mean equals 45, while the alternative hypothesis (H₁) asserts that the mean is not equal to 45 (two-tailed test).
Since the sample size is less than 30, a t-test is appropriate. The test statistic, t, is calculated as:
\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]
Where:
- \(\bar{x}\) = 50 (sample mean)
- \(\mu_0\) = 45 (hypothesized mean)
- \(s\) = 8 (sample standard deviation)
- \(n\) = 25 (sample size)
Calculating:
\[ t = \frac{50 - 45}{8 / \sqrt{25}} = \frac{5}{8 / 5} = \frac{5}{1.6} \approx 3.125 \]
Next, compare the absolute value of t (3.125) to the critical t-value at a 5% significance level for a two-tailed test with n−1 = 24 degrees of freedom. Consulting t-distribution tables, the critical value is approximately 2.064. Since 3.125 > 2.064, we reject the null hypothesis and conclude there is sufficient evidence at the 5% level to suggest the population mean is different from 45.
Exercise 2: Hypothesis Testing with Larger Sample Size (Using z-distribution)
Imagine Exercise 9.5.75 involves testing a population proportion with a sufficiently large sample size. For example, suppose a researcher claims that 60% of a population favors a new policy, based on a sample where 150 out of 250 individuals favor it. The null hypothesis is H₀: p = 0.60, and the alternative is H₁: p ≠ 0.60.
This is a two-tailed test for a proportion. The sample proportion is:
\[ \hat{p} = \frac{150}{250} = 0.60 \]
Since the sample size is large (n=250), a z-test is appropriate. The test statistic (z) is calculated as:
\[ z = \frac{\hat{p} - p_0}{\sqrt{p_0(1 - p_0)/n}} \]
Where:
- \(\hat{p}\) = 0.60
- \(p_0\) = 0.60
- \(n\) = 250
Calculating:
\[ z = \frac{0.60 - 0.60}{\sqrt{0.60 \times 0.40 / 250}} = 0 \]
Since the z-value is 0, it falls exactly at the critical point, indicating no evidence to reject the null hypothesis at the 5% significance level. Therefore, the data do not suggest a significant difference from the hypothesized proportion.
Conclusion
In hypothesis testing, the proper selection of test statistic (t or z) depends primarily on sample size and data type. For small samples (
References
- Illowsky, B., & Dean, S. (n.d.). Homework exercises for Hypothesis Testing (Chapter 9).
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