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Namequiz 2stat 244spring 2015week 31in A Large Population It

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Namequiz 2stat 244spring 2015week 31in A Large Population It Is

In a large population, it is known that a certain cold virus will cause the average body temperature to be 99.5 degrees. In a sample of 25 cold sufferers taking a new drug, the average temperature is 98.5 degrees, with a sample standard deviation of 1.0 degrees. Is there evidence that the drug is useful in lowering body temperature? Use α = 0.01.

In a recent poll, 55 out of 80 people surveyed approved of a certain land-use decision. Does this provide sufficient evidence to conclude that more than half of the population supports the decision? Use α = 0.05.

Paper For Above instruction

The aim of this paper is to analyze two statistical scenarios based on hypothesis testing to determine whether certain claims can be supported by data. The first scenario examines whether a new drug effectively lowers body temperature in individuals suffering from a cold virus, while the second assesses whether more than half of a population supports a specific land-use decision.

Scenario 1: Effectiveness of a New Cold Drug in Lowering Body Temperature

The null hypothesis (H■) posits that the drug has no effect, meaning the population mean body temperature remains at 99.5°F. Mathematically, H■: µ = 99.5. The alternative hypothesis (H■) asserts that the drug lowers body temperature, such that µ < 99.5°F.

Given the data: sample size (n) = 25, sample mean (x■) = 98.5°F, sample standard deviation (s) = 1.0°F, and significance level (α) = 0.01, we perform a one-sample t-test to evaluate the hypotheses. The t-statistic is calculated as:

t = (x■ - µ■) / (s / √n) = (98.5 - 99.5) / (1.0 / √25) = (-1.0) / (0.2) = -5.0

Using a t-distribution table or statistical software, the p-value corresponds to the probability of observing a t-value as extreme as -5.0 under the null hypothesis. For degrees of freedom (df) = 24, this p-value is very small, approximately less than 0.001.

Since p < α = 0.01, we reject the null hypothesis, providing strong evidence that the drug is effective in lowering body temperature in cold sufferers.

Scenario 2: Support for Land-Use Decision

The null hypothesis (H■) states that half or fewer people support the decision, H■: p ≤ 0.5. The

alternative hypothesis (H■) suggests that more than half support it, H■: p > 0.5.

From the sample, 55 out of 80 individuals support the decision, resulting in a sample proportion: p■ = 55 / 80 = 0.6875

The test statistic for proportions is: z = (p■ - p■) / √(p■(1 - p■) / n) = (0.6875 - 0.5) / √(0.5 * 0.5 / 80) ≈ 0.1875 / 0.0559 ≈ 3.355

Corresponding p-value for a one-tailed test (z > 3.355) is approximately 0.0004.

Since p < α = 0.05, we reject H■, concluding that there is sufficient evidence that more than half of the population supports the land-use decision.

Conclusion

Based on the hypothesis tests, the drug appears effective in reducing body temperature among cold sufferers, with strong statistical significance. Simultaneously, there is compelling evidence to support that a majority of the population favors the land-use decision. These conclusions are made with confidence levels set at 1% and 5% respectively, adhering to rigorous statistical standards.

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