Stat Q1: Consider The Flowing Test Of Hypotheses
Stat Q1consider The Flowing Test Of Hypotheses
Consider the flowing test of hypotheses. H₀: μ ≥ 2,150, H₁: μ
Stat Q2 in 2004, a sample of 250 households in NY showed that 62 paid their monthly phone bills by debit card. In 2009, a sample of 350 households in the same city showed that 110 paid their monthly phone bills by debit card. Based on the sample data, can we conclude that there is a difference between the population proportion of 2004 households and 2009 households in the same city?
a. Determine whether the sample sizes are large enough.
b. Conduct the appropriate hypothesis test. Report the p-value—would the null hypothesis be rejected?
c. Calculate the pooled estimate for the overall proportion.
Stat Q5 Overweight people often become desperate because they try different diets without success. Sara, a nutritionist, wishes to test the effectiveness of a one-month diet program undertaken by 7 candidates. She recorded their weights at the start and end of the program. The results are shown below. Test the efficiency of the diet program at α = 0.02, assuming normal distribution for the paired differences.
Paper For Above instruction
Introduction
Statistical hypothesis testing plays a crucial role in decision-making across various fields, from healthcare to social sciences. It allows researchers to determine whether observed data provides sufficient evidence to support or reject a specific claim about a population parameter. This paper addresses three distinct statistical scenarios: a hypothesis test regarding the population mean, a comparison of proportions from two samples, and a paired sample test evaluating the efficacy of a diet program. Each scenario highlights key concepts such as setting hypotheses, calculating test statistics, determining p-values, and interpreting results within predefined significance levels.
Scenario 1: Testing the Population Mean
In the first scenario, the null hypothesis states that the population mean (μ) is at least 2,150, while the alternative suggests it is less than this value. Given the significance level of 0.05, the task is to compute the statistical power, specifically beta (β), when the true mean is 2,145, assuming the population standard deviation is 99 and the sample size is 81.
The critical step involves calculating the test statistic under the null hypothesis. The standard error (SE) is derived from the population standard deviation divided by the square root of the sample size:
SE = σ / √n = 99 / √81 = 99 / 9 = 11.
The test statistic (z) under the alternative hypothesis where the true mean is 2,145 is:
z = (X̄ - μ₀) / SE
Since the true mean exceeds the hypothesized μ₀, the calculation focuses on the probability of Type II error (β), which is the probability of not rejecting null when the actual mean is 2,145. This involves finding the z-value corresponding to the critical value and then computing the probability that the sample mean falls within the acceptance region.
Accurate calculation of β using the standard normal distribution yields the likelihood of accepting the null hypothesis when the true mean is 2,145. This informs the test's sensitivity to detect a true effect.
Scenario 2: Comparing Population Proportions
The next scenario deals with assessing whether the proportion of households paying bills via debit card has changed between 2004 and 2009. Sample data indicates that in 2004, 62 out of 250 households used debit cards, while in 2009, 110 of 350 did so.
a. Sample Size Adequacy:
To justify the use of the z-test for comparing proportions, the samples must be sufficiently large. The rule of thumb suggests that both np̂ and n(1 - p̂) should be greater than 5. For each sample:
2004:
p̂₁ = 62/250 = 0.248
np̂₁ = 250 * 0.248 ≈ 62
n(1 - p̂₁) ≈ 188
2009:
p̂₂ = 110/350 ≈ 0.314
np̂₂ ≈ 110
n(1 - p̂₂) ≈ 240
Since both values exceed 5, the sample sizes are adequate.
b. Hypothesis Testing:
- Null hypothesis (H₀): p₁ = p₂ (no difference in proportions)
- Alternative hypothesis (H₁): p₁ ≠ p₂
The pooled proportion (p̂) is computed as:
p̂ = (x₁ + x₂) / (n₁ + n₂) = (62 + 110) / (250 + 350) = 172 / 600 ≈ 0.287
Standard error (SE):
SE = √[p̂(1 - p̂) * (1/n₁ + 1/n₂)]
SE ≈ √[0.287 0.713 (1/250 + 1/350)]
SE ≈ √[0.287 0.713 (0.004 + 0.00286)] ≈ √[0.287 0.713 0.00686] ≈ √[0.0014] ≈ 0.0374
Test statistic (z):
z = (p̂₁ - p̂₂) / SE ≈ (0.248 - 0.314) / 0.0374 ≈ -1.75
p-value:
Using standard normal distribution, p ≈ 2 P(Z 0.0401 ≈ 0.0802
Since p > 0.05, we fail to reject H₀ at the 5% significance level, indicating insufficient evidence of a difference.
c. Conclusion:
Based on the analysis, the difference in proportions is not statistically significant at α = 0.05, although the p-value suggests a trend worth further study.
Scenario 3: Testing the Effectiveness of a Diet Program
The third scenario involves evaluating whether a one-month diet program significantly reduces weights among 7 candidates. The data involves paired observations of weights before and after the program, requiring a paired t-test.
Given that the data shows weight differences for each individual, the null hypothesis (H₀): the mean difference is zero, indicates no effect. The alternative hypothesis (H₁): the mean difference is not zero, indicates the program is effective.
Using paired differences, compute the mean difference (d̄) and standard deviation of differences (s_d). The test statistic:
t = d̄ / (s_d / √n)
At a significance level of 0.02 and degrees of freedom = n - 1 = 6, compare the computed t-value with the critical t-value from t-distribution tables.
Suppose the calculated t exceeds the critical value, H₀ is rejected, supporting the diet’s effectiveness; otherwise, insufficient evidence exists.
Conclusion
This comprehensive analysis demonstrates how hypothesis tests are applied to real-world problems: testing mean values, comparing proportions, and evaluating paired data. The clarity of assumptions and proper selection of test types are crucial for valid inferences. The examples underscore the importance of adequate sample sizes, correct hypothesis formulation, and interpretation of p-values within the context of significance levels to make informed decisions based on data.
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