I Want To Test A Claim That The Mean Waste Generated By Adul

I Want To Test A Claim That Themeanwaste Generated By Adults In The

I want to test a claim that the mean waste generated by adults in the country is more than 3.5 pounds per person per day. In a random sample of 9 adults in the country, I find that the mean waste generated per person is 4.3 pounds per day with a standard deviation of 1.2 pounds. At the significance level of 0.01, I need to determine whether I can support the claim. The critical value provided is t = 2.897. I must find the standardized test statistic (t) to evaluate this claim.

Additionally, applying the Central Limit Theorem in another scenario, a friend wishes to know if men at his college are 72 inches tall, on average. He randomly sampled 10 men on his campus, measured their heights, and calculated the sample mean to be 69 inches. He concluded that men at his college do not differ significantly in height from Apex men because 69 inches is only 1.5 standard deviations less than 72 inches and therefore not statistically significant. This evaluation requires analysis of his procedure and conclusion.

Paper For Above instruction

To rigorously test the claim that the mean waste generated by adults exceeds 3.5 pounds per day, we employ hypothesis testing, specifically a t-test for a single mean. The null hypothesis (H₀) posits that the mean waste equals 3.5 pounds, while the alternative hypothesis (H₁) asserts that the mean waste is greater than 3.5 pounds. Formally, these are expressed as:

  • H₀: μ = 3.5
  • H₁: μ > 3.5

Given the sample size (n=9), which is less than 30, and assuming normality, the t-distribution models the sampling distribution accurately. The sample mean (x̄ = 4.3), sample standard deviation (s = 1.2), and the degrees of freedom (df = 8) form the basis for the test statistic. The formula for the t-statistic is:

t = (x̄ - μ₀) / (s / √n)

Replacing with the given values:

t = (4.3 - 3.5) / (1.2 / √9) = 0.8 / (1.2 / 3) = 0.8 / 0.4 = 2.0

The computed t-value of 2.0, against the critical t-value of 2.897 at a significance level of 0.01, suggests that the test statistic does not fall into the rejection region. Since 2.0

In the second scenario involving the height of men at a college, the friend approximates the significance of the difference using the z-score, calculated as 1.5 standard deviations less than 72 inches. This approach is generally acceptable when the population standard deviation (σ) is known; however, since only the sample standard deviation is given, a t-test should be employed for accuracy.

The sample mean (69 inches), sample size (n=10), and sample standard deviation (which is unspecified but presumed available) are used to formulate the test. The t-statistic for this scenario is:

t = (sample mean - hypothesized mean) / (sample standard deviation / √n)

In the absence of the sample standard deviation, the approximate z-score of 1.5 indicates that the observed mean is within a reasonable range of the hypothesized mean. Typically, a position 1.5 standard deviations away corresponds to a p-value around 0.13 (13%), which is above the common significance thresholds (such as 0.05). Thus, it is appropriate for the friend to conclude that there isn't sufficient evidence to assert a significant difference in heights between his sampled men and the population standard.

This evaluation highlights the importance of employing proper statistical tests aligned with the known parameters and sample data. Using a z-test without known population standard deviation is riskier; utilizing the t-test accommodates unknown σ and small sample sizes, providing more reliable results. The friend's conclusion, based on the raw standard deviations, is therefore reasonable, acknowledging the limitations of small samples and the need for proper statistical methods.

References

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