Show Work On Questions 8, 18, 38, 48, 6, And Everythi 055060

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

In this paper, I will analyze and present detailed solutions for selected statistical hypothesis testing problems drawn from a set of questions. The focus will be on questions 8-1, 8-3, 8-4, and 8-6, as specified. These questions cover a range of topics including t-tests, z-tests, confidence intervals, and hypothesis formulation, which are foundational in inferential statistics. The solutions will include step-by-step calculations, hypothesis statements, and interpretations based on data provided, fostering a clear understanding of the statistical methods employed.

Question 8-1

In the context of comparing the average marathon finishing times of women, the past average time was 4.62 hours. A recent sample of 2001 women recorded an average time of 4.29 hours with a standard deviation of 1.11 hours. The question asks whether this year's mean time is significantly different from the historical mean at a 5% significance level.

Null hypothesis (H0): μ = 4.62 hours

Alternative hypothesis (H1): μ ≠ 4.62 hours

Given data: sample mean (x̄) = 4.29, population standard deviation (σ) not directly given, but standard deviation from sample is 1.11. Since the sample size is large (n=2001), a z-test can be used.

Standard error (SE) = σ / √n = 1.11 / √2001 ≈ 1.11 / 44.73 ≈ 0.0248

Calculate z-statistic: z = (x̄ - μ0) / SE = (4.29 - 4.62) / 0.0248 ≈ -0.33 / 0.0248 ≈ -13.31

P-value (two-tailed) corresponding to z ≈ -13.31 is effectively 0 (p

Additionally, confidence interval at 95% (from the provided data) is (4.24, 4.34), which does not contain 4.62, supporting the rejection of H0. Therefore, we conclude that the average time for women to finish the marathon this year is significantly different from the past.

Question 8-3

This question involves testing whether the average amount of sugar in 5-pound sugar bags is at most 5 pounds. A random sample of 75 bags shows a mean of 4.95 lbs with a standard deviation of 0.5 lbs.

Null hypothesis: H0: μ ≤ 5

Alternative hypothesis: H1: μ > 5

Since the sample size is large, a z-test is appropriate. Calculate the standard error: SE = 0.5 / √75 ≈ 0.058

Compute z-value: z = (4.95 - 5) / 0.058 ≈ -0.05 / 0.058 ≈ -0.862

Using standard normal distribution tables, the p-value for z ≈ -0.86 (right-tailed test with H1: μ > 5) corresponds to 1 - P(Z

The conclusion aligns with support for the claim that the mean sugar content does not exceed 5 pounds, with the data not providing sufficient evidence to suggest otherwise.

Question 8-4

The hypothesis involves the proportion of students with loan debt exceeding $25,000. The sample size is 200, with 125 students reporting debt over $25,000.

Null hypothesis: H0: p ≥ 0.75

Alternative hypothesis: H1: p

Sample proportion: p̂ = 125/200 = 0.625

Standard error: SE = √[p0(1 - p0)/n] = √[0.75 * 0.25 / 200] ≈ √(0.1875 / 200) ≈ √0.0009375 ≈ 0.0306

Z-value: z = (p̂ - p0) / SE = (0.625 - 0.75) / 0.0306 ≈ -0.125 / 0.0306 ≈ -4.09

Using standard normal tables, the p-value is P(Z

Question 8-6

Testing whether the dropout rate exceeds 12%, the sample includes 500 students with 72 dropouts.

Null hypothesis: H0: p ≤ 0.12

Alternative hypothesis: H1: p > 0.12

Sample proportion: p̂ = 72/500 = 0.144

Standard error: SE = √[p0(1 - p0)/n] = √[0.12 * 0.88 / 500] ≈ √(0.1056 / 500) ≈ √0.0002112 ≈ 0.0145

Compute z: z = (0.144 - 0.12) / 0.0145 ≈ 0.024 / 0.0145 ≈ 1.65

The p-value for z = 1.65 is approximately 0.0495. Since p

In conclusion, the evidence suggests the dropout rate is significantly greater than 12%, which supports the original researcher’s suspicion.

References

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  • Depaoli, D., & van de Schoot, R. (2017). Bayesian Methods for the Social and Behavioral Sciences. Routledge.
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  • Ott, R. L., & Longnecker, M. (2010). An Introduction to Statistical Methods and Data Analysis (6th ed.). Brooks Cole.
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