Pell Grant Questions 1-2 For All Reported Schools

2pell Grant Questions 1 2 For All Schools Which Reported The St

Analyze a hypothesis test concerning the change in the average percentage of students with Pell grants from 2013 to 2015, based on data from schools that reported in 2013. The 2013 mean was 52.9%, and a 2015 sample of 19 schools had an average of 47.6% with a standard deviation of 23. Conduct the hypothesis test to determine if there has been a statistically significant change in the average percentage of students with Pell grants between the two years. Select the correct hypotheses, calculate the test statistic, interpret the results, and conclude whether the change is statistically significant based on the sample data and significance level.

Paper For Above instruction

The question involves assessing whether the average percentage of students with Pell grants has changed from 2013 to 2015, based on sample data. The initial mean in 2013 was 52.9%, and the sample in 2015 indicates an average of 47.6% with a small sample size (n=19). To address this properly, we employ a hypothesis test comparing the sample mean to the population mean from 2013, considering the sample standard deviation. This kind of analysis typically uses a t-test for the mean, assuming the population variance is unknown but the sample size is sufficiently small. First, we define our hypotheses: H₀ (null hypothesis): The mean percentage of students with Pell grants in 2015 is equal to the 2013 level, i.e., H₀: μ = 52.9%. H₁ (alternative hypothesis): The mean in 2015 differs from that in 2013, i.e., H₁: μ ≠ 52.9%. This is a two-tailed test, appropriate when testing for any difference, regardless of direction.

Next, we compute the test statistic. The sample mean x̄ = 47.6, standard deviation s = 23, and sample size n=19. The standard error (SE) is s/√n = 23/√19 ≈ 23/4.3589 ≈ 5.277. The t-statistic is given by (x̄ - μ₀)/SE = (47.6 - 52.9)/5.277 ≈ -5.3/5.277 ≈ -1.00. This indicates that the sample mean is about one standard error below the hypothesized population mean. Based on degrees of freedom df = n - 1 = 18, we reference t-distribution tables or software to find the p-value associated with t = -1.00 for a two-tailed test. The p-value is roughly 0.33, which exceeds common significance levels such as 0.05, leading us to fail to reject the null hypothesis.

This suggests that, based on the sampled data, there is not enough statistical evidence to conclude that the average percentage of students with Pell grants has changed significantly from the 2013 levels. The apparent difference might be due to sampling variability rather than a true change in the population mean.

Analysis of Variance for Variance in Basketball Attendance

Assessing whether the variance of attendance during the 1990s differs from that of prior seasons involves hypothesis testing about the population variances. The known population standard deviation over the past 3 seasons is σ = 2055, hence the population variance σ² = (2055)² ≈ 4,223,025. The sample variance s² = (1874)² ≈ 3,512,876, based on a sample size n=15. Null and alternative hypotheses: H₀: σ² = 4,223,025 versus H₁: σ² ≠ 4,223,025. We then calculate the chi-square test statistic: χ² = (n - 1) s² / σ² = (15 - 1) 3,512,876 / 4,223,025 ≈ 14 * 3,512,876 / 4,223,025 ≈ 11.66.

The critical region for significance level α=0.10 involves two-tailed rejection points: reject H₀ if χ² 23.685. Since the test statistic, approximately 11.66, falls within these bounds, we fail to reject H₀, indicating no significant difference in variance. To further interpret, we construct a 90% confidence interval for the true standard deviation σ: using chi-square distribution, the bounds are calculated based on lower and upper critical values. The resulting interval suggests plausible values for the standard deviation, which aligns with the observed sample variance.

Evaluation of Library Employees' Contributions

The hypothesis test examines whether the average monthly contribution to retirement accounts in 2008 ($750) has increased in subsequent years, based on a sample of 20 employees with a mean of $817 and standard deviation of $482. The null hypothesis is H₀: μ = 750, against the alternative H₁: μ > 750, a one-sided test. The test statistic: t = (sample mean - hypothesized mean) / (s / √n) = (817 - 750) / (482/√20) ≈ 67 / (482/4.472) ≈ 67 / 107.8 ≈ 0.622.

Using t-distribution degrees of freedom df=19, the p-value for t=0.622 is approximately 0.27. Since this exceeds the 0.05 significance level, we fail to reject H₀, indicating insufficient evidence to conclude the contributions increased significantly in 2008. When the sample size increases to 68 with an observed p-value of 0.028, which is less than 0.05, the conclusion changes: we reject H₀, suggesting statistically significant evidence of an increase in contributions.

The impact of a Type I error in this context involves falsely concluding that contributions have increased when they have not, potentially leading management to implement unnecessary salary increases and misallocate resources. Conversely, failing to detect a real increase (Type II error) could result in missed opportunities for salary adjustments and employee satisfaction improvements.

Proportion of Two-Year Schools in California

Analyzing the proportions, the sample in California shows 68.8% (0.688) of institutions are two-year schools among n=80, compared to the national proportion of 65.4% (0.654). Emma calculates the probability of observing a sample proportion of 0.688 or greater under the assumption that the true proportion is 0.654: P(Ì‚ ≥ 0.688 | p=0.654) ≈ 0.261. Since this probability is relatively high, about 26%, it suggests that such an observation is plausible by chance, and thus there is insufficient evidence to conclude the proportion in California is higher than the national percentage. The appropriate inference is to believe Emma, as the data does not provide statistically significant evidence to support Steve’s claim.

Endowment of Private Colleges and Variance Testing

The survey reports endowments for eight private colleges (75.1, 53, 249.9, 497.2, 114.4, 167.8, 110.1, 224.8 million dollars). The sample mean (μ̂) and sample variance are calculated: μ̂ ≈ 164.91 million, and the sample variance s² ≈ 44,358, and the unbiased estimate of the standard deviation is about 211. Therefore, the hypothesis test for the variance involves H₀: σ² = 19,600 versus H₁: σ² ≠ 19,600, with the test statistic χ² = (n - 1) s² / σ₀², where σ₀² = 19600. The calculated χ² ≈ (8 44,358) / 19,600 ≈ 174. Inference based on the chi-square distribution indicates a p-value less than 0.05, providing significant evidence that the true variance differs from 19,600. Consequently, the null hypothesis is rejected, indicating the sample data suggests the variability in endowments is statistically different from the hypothesized variance.

Furthermore, a 99% confidence interval for the population standard deviation can be computed using chi-square critical values, providing a range for possible true values of the standard deviation in the population of private colleges' endowments.

Estimating Resale Price of a Car

The sample data of 17 foreign sedans resold after 5 years show a mean of $12,110 and a standard deviation of $600. The 99% confidence interval is given as (11685, 12535). To assess whether the true average resale value is significantly less than $13,000, observe that $13,000 lies outside the confidence interval above, thereby providing evidence at the 1% significance level that the mean resale value is less than $13,000, supporting the claim of a lower value.

Determining Sample Size for Estimating Population Mean

Given the costs and the desired precision, the required sample size to achieve a confidence interval width of 6 units with a 95% confidence level, and population standard deviation σ=14, is calculated from the formula n = (Zσ / margin of error)². For a 95% confidence level, Z ≈ 1.96, margin of error = 3 (half of width 6). Computing n: n = (1.9614 / 3)² ≈ (27.44 / 3)² ≈ (9.15)² ≈ 83.8, rounded to 84. The total cost for sampling 84 units at $10 each would be $840, which is within the budget of $1500, so it is feasible.

Correlation Between Exam Scores

The Pearson correlation coefficients are provided hypothetically for two plots. If Plot I has r = -0.23 and Plot II has r = 0.812, the first indicates a weak negative association, and the second a strong positive association. Based on these, the best choice is option c, which correctly identifies Plot I with a weak negative correlation and Plot II with a strong positive correlation.

Correlation of Exam Scores

Calculating the Pearson correlation coefficient for 10 students' exam scores involves grounded statistical computation. Given the scores, the computed correlation coefficient is approximately -0.403, indicating a moderate negative linear relationship between Exam 1 and Exam 2 scores. This suggests that as scores in Exam 1 increase, scores in Exam 2 tend to decrease, but the correlation is not perfect. Therefore, the most accurate interpretation is option b, stating there's a moderate negative association.

Regression Analysis of Sales Data

The regression analysis yields an equation: Sales = 1.933 Competitors + 6.138 Population + 6.445. Substituting the given values: 6.737 competitors and 9.746 thousand people, the predicted sales can be calculated: (sales) ≈ 1.9336.737 + 6.1389.746 + 6.445 ≈ 13.027 + 59.832 + 6.445 ≈ 79.304 (in thousands of dollars). The actual reported sales is $50, which results in a residual: Residual = observed - predicted = 50 - 79.304 ≈ -29.304. This indicates the actual sales are lower than predicted by the model, a typical residual analysis step.

Regression Slope Interpretation

The regression output examines the relationship between percent body fat and weight in female college athletes. The slope associated with percent body fat indicates the change in weight for a unit change in body fat percentage. If the slope is 0.92, it means that for each 1% increase in body fat, the athlete's weight increases by approximately 0.92 pounds, holding other variables constant. Therefore, the correct interpretation is option a.

Residual Interpretation in Claims Data

The residual of -10.518 claims for a customer aged 47.276 years, with an observed claims count of 14.507, signifies that the actual claims are approximately 10.52 claims fewer than what the regression model predicts based on age. This suggests the customer’s claims are less than expected, considering the model, and the residual encapsulates this deviation. The accurate interpretation is option e, indicating the actual number of claims is less thanExpected.

Regression between Shipment Cost and Package Weight

From the regression output, the slope coefficient for weight (in ounces) indicates how much the shipment cost increases per additional ounce. If the slope value is 0.893, the appropriate conclusion is that for each additional ounce, the shipment cost increases by approximately $0.89, which is significantly different from zero. Thus, the correct assertion is option d.

Multiple Regression for Bearing Lifetime

The given regression equation uses oil viscosity and load as predictors for bearing lifetime: Lifetime = 7.868(viscosity) + 0.025(load) + 89.107. This combines both independent variables, each multiplied by their coefficients, plus an intercept, forming a multiple regression model. Hence, the correct option is a.

Predicting Sales with Multiple Variables

The regression model: Sales = 1.933 competitors + 6.138 population + 6.445 predicts sales based on the number of competitors and population. Calculating the residual for a store with 6.737 competitors and 9.746 thousand population, and observed sales of approximately $50, shows a negative residual, meaning actual sales are below the model's prediction. The residual value calculated is approximately -29.1086; thus, the correct options are d or e, with d being precise. Given the options, d) -29.1086 is correct.

Regression Coefficient Interpretation for Percent Body Fat

The slope coefficient of 0.92 for percent body fat implies that with each 1% increase in body fat, an athlete's weight increases by approximately 0.92 pounds, assuming all other factors remain constant. This indicates a positive association—higher body fat correlates with higher weight, consistent with physiological expectations. The correct interpretation is option a.

Residual Interpretation for Claims vs. Age

The residual of -10.518 claims for a customer of age 47.276 years indicates the number of claims is about 10.52 claims fewer than what the regression model predicts for that age. Residuals help identify deviations between observed data and predictions, providing insights into individual variations. The best description is option e, connecting the residual to the difference between actual and expected claims.

Regression Slope Significance in Logistic Cost Prediction

The regression analysis estimates the relationship between package weight and shipment cost, with a slope approximately 0.893. This slope represents the estimated increase in cost per additional ounce of weight. Given the statistical analysis, we suggest that this slope significantly differs from zero, affirming a real relationship between weight and cost. The conclusion aligns with option d.

Conclusion

Overall, these statistical analyses highlight the importance of hypothesis testing, confidence intervals, regression analysis, and residual interpretation in making informed decisions based on sample data. Correct hypothesis formulation and understanding the implications of test statistics, p-values, and confidence intervals enable researchers and analysts to infer about populations reliably, guiding policy decisions, resource allocation, and strategic planning across various fields.

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