EGCP 401 Spring 2015 Homework 4 Due February 9, 2015

Egcp 401 Spring 2015homework 4 Due February 9, 20152 8 2 9 2 52

Compare annual operating costs of two rides: Tummy Tugger and Head Buzzer, given fixed and variable costs; find the breakeven point in visitors; create a graph illustrating total costs and preferred alternatives; analyze a breakeven graph for investment; determine total revenue and costs, breakeven level, profit/loss at 1500 units, and costs at that level; calculate interest rate on a past investment; conduct confidence interval estimate for credit card balances; formulate hypotheses and compute test statistic and p-value for a phantom's mineral density measurement; perform a t-test to compare differences in mpg calculations; test if nicotine content exceeds a claimed mean; design a study comparing website design effectiveness; analyze data on spending behavior under sad and neutral conditions; examine last effects of horror movies on bedtime and waking life symptoms.

Sample Paper For Above instruction

The comparison of operational costs between the Tummy Tugger and the Head Buzzer rides provides crucial insight for an amusement park aiming to optimize its expenditures. To determine the point at which the costs of both rides are equal, we formulate a mathematical model that includes their fixed and variable costs. The Tummy Tugger incurs a fixed cost of $10,000 annually with a variable cost of $2.50 per visitor, while the Head Buzzer has a fixed cost of $4,000 with a variable cost of $4 per visitor. Let x represent the number of visitors.

The total annual cost for the Tummy Tugger (T) is given by T = 10,000 + 2.50x, and for the Head Buzzer (H) it is H = 4,000 + 4x. To find the breakeven point, set T = H: 10,000 + 2.50x = 4,000 + 4x. Subtract 4,000 from both sides: 6,000 + 2.50x = 4x. Subtract 2.50x from both sides: 6,000 = 1.50x. Solving for x gives x = 6,000 / 1.50 = 4,000 visitors. Therefore, at 4,000 visitors, the costs associated with both rides are equal.

Next, we develop a graph to visualize the total costs for each ride, the breakeven point, and the ranges where each ride is more cost-effective. The total cost lines for the two rides are expressed as: T = 10,000 + 2.50x for the Tummy Tugger and H = 4,000 + 4x for the Head Buzzer. Plotting these equations on a coordinate plane, with visitors (x) on the x-axis and total costs on the y-axis, reveals the point of intersection at x=4000. To the left of this point, the Head Buzzer's costs are lower, indicating it is more economical for lower visitor counts. Beyond this point, the Tummy Tugger becomes more cost-effective as the total cost line for it falls below that of the Head Buzzer.

Considering the breakeven analysis, amusement park managers should choose the ride based on expected visitor volumes. For fewer than 4,000 visitors annually, the Head Buzzer minimizes costs; for more than 4,000 visitors, the Tummy Tugger offers savings. This analysis aids in strategic decisions regarding ride investments, especially as visitor estimates fluctuate in seasonal or promotional periods.

In investment analysis, a breakeven graph might depict the total revenue and total cost lines for a certain project. If, for example, a firm invests in a new technology ten years ago when the investment was $450,000, now worth $1,000,000, the simple interest rate can be calculated using the formula: interest rate = (Final amount - Principal) / (Principal × Time). Substituting the values yields interest rate = ($1,000,000 - $450,000) / ($450,000 × 10) = $550,000 / $4,500,000 ≈ 0.1222 or 12.22% per annum.

Confidence intervals are statistical tools used to estimate the true mean of a population parameter. For example, analyzing credit card balances among 1200 loan applicants, with an average of $3173, standard deviation $3500, and a confidence level of 95%, the confidence interval is computed as: mean ± z*(standard deviation / √n). For a 95% level, z ≈ 1.96. Therefore, the interval is $3173 ± 1.96 × ($3500 / √1200) ≈ $3173 ± 1.96 × $1012. The resulting interval estimates where the true mean lies with 95% confidence.

When assessing the accuracy of a device measuring mineral density via a phantom, null and alternative hypotheses are constructed: H0: μ = 1.4 (the known mineral density), Ha: μ ≠ 1.4 (the measured mean differs). Conducting a significance test involves calculating the test statistic, typically a z-score, and comparing it to critical values or computing a p-value. For example, with 10 scans, a sample mean of 1.5, and standard deviation known, the z-score is calculated as (sample mean - hypothesized mean) / (standard deviation / √n). Based on the p-value derived, one can decide whether to reject H0.

The comparison of two sets of data, such as the differences in miles per gallon calculations, involves a t-test for the difference in means. State the null hypothesis as no difference between the two means, and the alternative as a difference exists. Calculate the t-statistic based on sample means, variances, and sizes. For instance, with a t-value of 2.45 at a significance level of 0.05 and degrees of freedom approximately 18 (using the Welch-Satterthwaite equation), a p-value less than 0.05 leads to rejecting H0, indicating a significant difference.

In hypothesis testing for nicotine content, setting the null hypothesis as the mean nicotine content being 0.9 mg, and the alternative as exceeding 0.9 mg, the test statistic z=1.83. At a significance level of 5%, since z=1.83 is greater than the critical z of 1.645, the increase is statistically significant. At the 1% level, the critical z is 2.33, thus the result is not significant at this stricter level.

Designing a comparative study between two website designs involves deciding on the appropriate test type. If the goal is to detect any difference, a two-sided test is appropriate. Degrees of freedom in a t-test with two independent samples are calculated via the Welch approximation. For a significance level of 0.05, the critical t-value depends on the degrees of freedom. For 15 days in each group (total 30), the degrees of freedom approximate to 28. A t-value of 2.45 exceeds the critical value (~2.048), indicating significant difference, and the p-value would be below 0.05, leading to rejection of the null hypothesis.

Analyzing behavioral data from a study involving sadness and spending reveals that the null hypothesis posits no difference in purchase prices between groups. Conducting a t-test requires calculating the mean and standard deviation for each group, then computing the t-statistic based on these values and sample size. Comparing the t-statistic to critical thresholds or calculating the p-value enables conclusion: if p

Finally, assessing the association between simultaneous symptoms reported in a study involves calculating the percentage of students with waking-life symptoms. The analysis of contingency tables via Chi-square tests determines whether the presence of waking symptoms is independent of bedtime symptoms. Null hypothesis states independence; a significant chi-square statistic (p

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