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This worksheet contains values required for MegaStat charts, including boxplots and dotplots. It also includes data related to Lincolnville School District buses, such as manufacturer, engine type, capacity, maintenance cost, age, odometer miles, and miles since last maintenance. Additionally, data from Century National Bank includes customer checking accounts, ATM transactions, bank services, and debit card usage. The assignment involves analyzing these data sets to answer specific statistical and business questions, including calculating probabilities, confidence intervals, and conducting data comparisons.

Sample Paper For Above instruction

The provided data sets from the Lincolnville School District buses and Century National Bank offer rich opportunities for statistical analysis, allowing us to explore diverse questions related to manufacturing costs, maintenance, banking behaviors, and polling data. This paper systematically addresses each of the assigned problems, applying relevant statistical concepts such as standard error, probability under the normal distribution, confidence intervals, and hypothesis testing, supported by proper data analysis techniques.

Problem 44: Milk Price Experiment Analysis

The first problem involves evaluating the average price of milk as an indicator of inflation. Given that the population mean price per gallon is $3.82 with a standard deviation of $0.33, and a sample size of 40 stores, we first calculate the standard error of the mean (SEM). The SEM is determined by dividing the standard deviation by the square root of the sample size:

SEM = 0.33 / √40 ≈ 0.0522

This indicates the amount of variability expected in the sample mean if we repeated the sampling process multiple times.

Next, we compute the probability that the sample mean lies between $3.78 and $3.86. Converting these to standard normal z-scores:

z = (X - μ) / σ/sqrt(n)

For $3.78: z = (3.78 - 3.82) / 0.0522 ≈ -0.766

For $3.86: z = (3.86 - 3.82) / 0.0522 ≈ 0.766

Using standard normal tables, the probability that z falls between -0.766 and 0.766 is approximately 0.45, indicating a 45% chance that the sample mean falls within this range.

Then, the probability that the difference between the sample mean and the population mean is less than $0.01 involves calculating the probability that the sample mean is between $3.81 and $3.83:

z for $3.81: (3.81 - 3.82) / 0.0522 ≈ -0.019

z for $3.83: (3.83 - 3.82) / 0.0522 ≈ 0.019

Very little probability, approximately 0.02, indicates that such a small difference is unlikely due to random variation.

Finally, the probability that the sample mean exceeds $3.92 is calculated by:

z = (3.92 - 3.82) / 0.0522 ≈ 1.916

From the standard normal distribution table, this corresponds to approximately 0.03, indicating a 3% chance that the sample mean is greater than $3.92.

Problem 48: Lincolnville School District Bus Data Analysis

The data from Lincolnville reveal the mean maintenance cost per year reported by manufacturers as $4,400 with a standard deviation of $1,000. To assess whether the Lincolnville buses align with this figure, we calculate the sample mean from the data and compare it with the established mean. Suppose the Lincolnville sample has a mean maintenance cost of $4,200 based on, for example, 10 buses. The probability of observing such a mean or higher, assuming the population mean is $4,400, is computed using the z-score:

z = (X̄ - μ) / (σ / √n) = (4200 - 4400) / (1000 / √10) ≈ -0.632

According to standard normal distribution, this corresponds to a probability of approximately 0.262, indicating that Lincolnville's data does not significantly deviate from the manufacturer's expectation.

This suggests that Lincolnville's maintenance costs are reasonably consistent with manufacturer reports, supporting efficiency in maintenance management.

Problem 56: Health Insurance Premium Confidence Interval

The sample of 20 subscribers reports an average annual premium of $10,979 with a standard deviation of $1,000. To develop a 90% confidence interval for the population mean, we use t-distribution owing to the small sample size:

Degrees of Freedom = 19; t-value ≈ 1.734

The confidence interval is given by:

CI = X̄ ± t (s / √n) = 10979 ± 1.734 (1000 / √20) ≈ 10979 ± 1.734 * 223.607 ≈ 10979 ± 387.7

Thus, the 90% confidence interval is approximately ($10,591.30, $11,366.70).

To determine the required sample size to estimate the population mean within $250 at 99% confidence, we rearrange the sample size formula:

n = (Z * s / E)^2

where Z ≈ 2.576 for 99% confidence and E = 250:

n = (2.576 * 1000 / 250)^2 ≈ (10.304)^2 ≈ 106.2

We round up to 107, indicating that a sample of 107 subscribers is needed for the desired precision.

Problem 66: Election Poll Confidence Intervals and Probabilities

The poll reports 52% for the Republican candidate based on 1,000 voters. The standard error for the proportion is:

SE = √[p(1-p) / n] = √[0.52 * 0.48 / 1000] ≈ 0.0157

The 95% confidence interval using z ≈ 1.96 is:

CI = 0.52 ± 1.96 * 0.0157 ≈ (0.489, 0.551)

This suggests that we are 95% confident the true proportion falls within approximately 48.9% and 55.1%.

Estimating the probability that the Democratic candidate is actually leading involves interpreting the confidence interval. Since the interval does not include 0.5, the probability that the Democratic candidate leads is less than the lower bound of the interval.

Repeating the process with 3,000 voters, the standard error reduces to:

SE = √[0.52 * 0.48 / 3000] ≈ 0.009

The confidence interval tightens to:

0.52 ± 1.96 * 0.009 ≈ (0.502, 0.538)

implying increased confidence in the estimate of the candidate's support.

Problem 71: Lincolnville Bus Maintenance and Odometer Miles

Using the Lincolnville bus data, the goal is to construct 95% confidence intervals for the mean maintenance cost and odometer miles. Suppose the sample mean maintenance cost is $4,200 with a standard deviation of $800 and a sample size of 15 buses:

CI for mean maintenance cost:

t-value with 14 df at 95% ≈ 2.145

CI = 4200 ± 2.145 (800 / √15) ≈ 4200 ± 2.145 206.2 ≈ 4200 ± 442

Resulting in a range of approximately ($3758, $4642).

Similarly, if the average odometer miles are 70,000 with a standard deviation of 10,000:

CI for odometer miles:

t-value ≈ 2.145

CI = 70000 ± 2.145 (10000 / √15) ≈ 70000 ± 2.145 258.2 ≈ 70000 ± 555

Indicating that the average odometer miles likely lie within approximately (69,445, 70,555).

A memo to the transportation official would summarize these findings, emphasizing the cost and mileage ranges, and comparing them to standard benchmarks or manufacturer data.

Analysis of Bank Customer Data and Debit Card Usage

For the Century National Bank data, the analysis aims to estimate the proportion of customers using debit cards through constructing a confidence interval for the population proportion. Suppose out of 60 sampled customers, 36 use debit cards, giving a sample proportion of 0.6. The standard error is:

SE = √[p(1-p) / n] = √[0.6 * 0.4 / 60] ≈ 0.063

The 95% confidence interval for the true proportion is:

0.6 ± 1.96 * 0.063 ≈ (0.476, 0.724)

Since this interval lies entirely above 0.5, it is reasonable to conclude that more than half of the customers use debit cards.

Such insights inform the bank’s strategic planning regarding debit card promotion and resource allocation across branches in different cities.

In conclusion, the comprehensive analysis of bus maintenance, banking data, opinions polls, and inflation measures illustrates the profound utility of statistical methods in real-world decision-making. Confidence intervals provide valuable estimates of population parameters, probabilities aid in risk assessment, and descriptive statistics reveal underlying data patterns, supporting organizations in optimizing operations.

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