Show Work On Questions 8, 18, 38, 48, 6, And Everything Else
Show Work On Questions 8 18 38 48 6 Everything Else Work Does Not
Show work on questions 8-1,8-3,8-4,8-6, everything else work does not need to be shown.
Paper For Above instruction
This paper addresses several hypothesis testing problems involving means, proportions, confidence intervals, correlation, and regression analysis. Each problem is analyzed systematically, providing step-by-step solutions including formulation of hypotheses, selection of appropriate tests, calculation of test statistics, p-values or critical values, and interpretation of results in context. The discussion emphasizes understanding the statistical methodology and correctly applying formulas for t-tests, z-tests, confidence intervals, and correlation coefficients, as well as chi-square tests where applicable.
Introduction
Hypothesis testing is an essential statistical tool used to make inferences about population parameters based on sample data. Correctly formulating null and alternative hypotheses is crucial, as is choosing the appropriate test statistic, whether z-test, t-test, or chi-square test, depending on data characteristics such as sample size, population variance knowledge, and data type. In this essay, various problems from educational and medical research contexts illustrate key concepts in hypothesis testing, confidence intervals, and correlation analysis.
Question 8-1: Comparing Marathon Finish Times
The question examines whether the average finishing time for women in the San Francisco Marathon has significantly decreased compared to past years. The null hypothesis (H₀) states that the population mean remains 4.62 hours, while the alternative suggests a difference exists. Given a sample mean of 4.29 hours, standard deviation of 1.11 hours, and sample size of 200, a z-test for a population mean is appropriate due to the large sample size. The z-statistic is calculated as:
z = (sample mean - population mean) / (standard deviation / √n) = (4.29 - 4.62) / (1.11 / √200) ≈ -2.86
Using standard normal distribution tables, the p-value associated with z ≈ -2.86 (two-sided test) is approximately 0.0042, which is less than α = 0.05. This indicates sufficient evidence to reject H₀ and conclude that the average marathon time has significantly changed. The options in the question reflect various interpretations, but the correct conclusion is that a significant difference exists, supporting the claim that times are different from past records.
Question 8-2: Cost to Rehabilitate Football Players
The null hypothesis posits that the population mean cost is at least $28,500 (H₀: μ ≥ 28,500). The alternative suggests it is less (
t = (30,885 - 28,500) / (1,123 / √15) ≈ 4.53
Degrees of freedom = 14. Looking up the p-value for t ≈ 4.53 with df=14 yields a p-value close to 0.0004, which is less than the significance level (not specified but typically 0.05). Therefore, there is strong evidence to reject the null hypothesis, implying the average rehabilitation cost exceeds the claim of $28,500.
Question 8-3: Sugar in Packaging
The food inspector tests whether the mean sugar content in 5-pound bags is at most 5 pounds. Null hypothesis: H₀: μ ≤ 5; alternative: H₁: μ > 5. The sample mean is 4.95 lbs, sample size 75, standard deviation 0.5 lbs. Since the sample size is large, a z-test applies:
z = (4.95 - 5) / (0.5 / √75) ≈ -1.83
Because the alternative hypothesis is right-tailed (μ > 5), the p-value is P(z > -1.83) ≈ 0.9664. The p-value indicates support for the null hypothesis; thus, the sample does not provide sufficient evidence to conclude the mean exceeds 5 pounds, consistent with packaging standards.
Question 8-4: Percent of Graduating Seniors with Student Loan Debt
Null hypothesis: p ≥ 0.75; alternative: p
z = (0.625 - 0.75) / √[0.75 * 0.25 / 200] ≈ -3.09
Corresponding p-value ≈ 0.001, which is less than α = 0.01. Thus, we reject H₀ and conclude that the proportion of students with debt exceeding $25,000 is less than 75%, indicating a significant decrease.
Question 8-5: Oxygen Uptake in Athletes
The null hypothesis claims that athletes’ maximum oxygen uptake is equal to the average of all adults (μ = 39). Alternative: μ > 39. Sample mean = 42.3, standard deviation = 6, sample size = 15. A t-test applies:
t = (42.3 - 39) / (6 / √15) ≈ 2.36
Degrees of freedom = 14. The p-value for t = 2.36 with df=14 is approximately 0.034. Since α = 0.02, p > α; thus, there is not enough evidence to support the claim that athletes’ maximum oxygen uptake exceeds 39 at the 2% significance level.
Question 8-6: Dropout Rate in California
Null hypothesis: p = 0.12; alternative: p > 0.12. The sample proportion is 72/500 = 0.144. The z-statistic is:
z = (0.144 - 0.12) / √[0.12 * 0.88 / 500] ≈ 2.05
Corresponding p-value ≈ 0.0202. At α = 0.05, since p ≤ α, we reject H₀, supporting the claim that the dropout rate exceeds 12%.
Question 8-8: Salary of Part-Time Instructors
Null hypothesis: μ ≤ $45,000; alternative: μ > $45,000. The sample mean = $42,500, sample std dev = $3,100, sample size = 25. The t-statistic is:
t = (42,500 - 45,000) / (3,100 / √25) ≈ -3.23
Since the alternative hypothesis tests for μ > $45,000, and the t-value is negative, the evidence suggests the mean is less, not greater, so we do not reject H₀. The appropriate test is a t-test (b). The critical value for α=0.10 (one-tailed) with df=24 is approximately 1.318.
Question 8-10: P-value for L.A. Lifeguards
Sample size = 50; sample mean = 29; standard deviation = 2.5; hypothesized mean = 28. The t-statistic is:
t = (29 - 28) / (2.5 / √50) ≈ 2.83
Degrees of freedom = 49. Using a t-distribution table, p-value (two-sided) for t=2.83 ≈ 0.006. Rounded to three decimal places, the p-value is approximately 0.006.
Question 8-11: P-value in a Two-tail T-Test
Given t = -1.523 with n=13, degrees of freedom = 12. Using t-distribution tables or calculator, the p-value for two-tailed test is approximately between 0.095 and 0.125, meaning:
-1.52
Question 8-12: Critical Value for Left-tailed Test
At α = 0.02 and large enough sample size, the critical z-value (since normal distribution is used) for a left-tail test is approximately -2.054.
Answer: -2.054, -2.054
Conclusion
This analysis demonstrates the application of hypothesis testing principles, from formulating hypotheses to interpreting p-values and critical values across various contexts. Accurate calculations, understanding the direction of tests, and proper interpretation are critical for valid statistical inferences in research, public health, education, and social sciences.
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