Ch 8 Problem 44: An Economist Uses The Price Of A Gallon Of

Ch 8problem 44an Economist Uses The Price Of A Gallon Of Milk As A Mea

Ch 8problem 44an Economist Uses The Price Of A Gallon Of Milk As A Mea

Analyze the given scenario involving the price of a gallon of milk as a measure of inflation, including calculations of standard error, probabilities related to sample means, and comparisons to population parameters. Also, interpret sample data from school buses, survey health insurance premiums, opinion polls, and banking customer data, applying confidence intervals and hypothesis testing where appropriate.

Paper For Above instruction

The analysis of economic indicators, such as the price of milk, alongside statistical inference techniques, provides essential insights into market behavior, inflation, and consumer trends. This paper explores multiple statistical problems related to such indicators: calculating standard error, determining probabilities of sample means, developing confidence intervals, and evaluating hypotheses, drawing on real-world data including milk prices, bus maintenance costs, health insurance premiums, election polls, and banking customer behaviors.

Statistical Analysis of Milk Price and Inflation Measures

An economist utilizes the average price of a gallon of milk as a proxy for measuring inflation. The given mean price is $3.82, with a population standard deviation of $0.33, based on data from numerous stores. To estimate the precision of this mean, the standard error (SE) of the mean must be calculated. The SE formula is:

SE = σ / √n

where σ is the population standard deviation, and n is the sample size. With σ = 0.33 and n = 40 (store samples), the calculation yields:

SE = 0.33 / √40 ≈ 0.33 / 6.3246 ≈ 0.0522

Therefore, the standard error of the mean milk price in this experiment is approximately $0.0522.

Next, calculating the probability that the sample mean falls between $3.78 and $3.86 involves standardizing these bounds into z-scores:

z = (X̄ - μ) / SE

For the lower bound ($3.78):

z₁ = (3.78 - 3.82) / 0.0522 ≈ -0.038 / 0.0522 ≈ -0.727

For the upper bound ($3.86):

z₂ = (3.86 - 3.82) / 0.0522 ≈ 0.038 / 0.0522 ≈ 0.727

Using standard normal distribution tables or software, the probability that Z is between -0.727 and 0.727 is approximately:

P(-0.727  Φ(0.727) - 1 ≈ 2  0.7664 - 1 ≈ 0.5328

Hence, there is roughly a 53.28% chance that the average milk price obtained from such a sample falls within the specified bounds.

The third probability involves the likelihood that the difference between the sample mean and the population mean is less than $0.01. Since the sampling distribution of the mean is approximately normal, this corresponds to:

P(|X̄ - μ|

Standardizing:

z = 0.01 / SE ≈ 0.01 / 0.0522 ≈ 0.1915

Therefore, this probability is:

P(-0.1915  Φ(0.1915) - 1 ≈ 2  0.5764 - 1 ≈ 0.1528

Indicating a roughly 15.28% chance that the sample mean deviates from the true mean by less than a cent, reflecting the variability inherent in sampling.

Analysis of School Bus Maintenance Costs

Based on manufacturer data, the average annual maintenance cost per school bus is $4,400, with a standard deviation of $1,000. The Lincolnville School District buses' data can be compared to this benchmark via confidence intervals. To estimate the mean maintenance cost, the sample mean (not provided explicitly) would be used, but assuming it aligns with the manufacturer’s claim, the confidence interval can be constructed using the formula:

CI = x̄ ± z*(σ/√n)

With a large sample (n=, for example, say 30 buses), the margin of error for a 95% confidence level (z ≈ 1.96) is:

ME = 1.96 * (1000/√n)

Suppose the sample mean is calculated; then the interval can be checked against $4,400. If the interval includes this value, the district’s costs are in line with manufacturer data; otherwise, they may differ significantly.

Calculating the probability that Lincolnville's actual mean maintenance cost exceeds the manufacturer's reported mean involves standard normal calculations. If the district's sample mean is, for example, $4,600, the z-score would be:

z = (4600 - 4400) / 1000 = 0.2

Using standard normal probabilities, the probability that the true mean is greater than $4,400, given the sample mean, is:

P(μ > 4400) = 1 - Φ(z) ≈ 1 - 0.5793 ≈ 0.4207

This indicates about a 42% chance that the true mean exceeds the reported $4,400, based on current data, which could inform budgetary and maintenance planning decisions.

Analysis of Health Insurance Premium Survey

The survey of 20 families reports a mean annual premium of $10,979 with a sample standard deviation of $1,000. The 90% confidence interval for the population mean can be calculated using the t-distribution (degrees of freedom = 19). The critical t-value for 90% confidence and 19 df is approximately 1.729. The margin of error is:

ME = t  (s/√n) = 1.729  (1000/√20) ≈ 1.729  223.6 ≈ 386.3

Thus, the confidence interval is:

(10979 - 386.3, 10979 + 386.3) ≈ ($10,592.7, $11,365.3)

There is 90% confidence that the true average annual premium for all families falls within this range.

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

n = (z  s / E)^2

Using z* for 99% confidence (approximately 2.576):

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

Rounding up, a sample of at least 107 families would be necessary to achieve this precision.

Opinion Polls and Confidence in Political Support

The poll of 1,000 voters indicates that 52% favor the Republican candidate. The confidence interval for this proportion at 95% confidence uses the formula:

CI = p̂ ± z  √[p̂(1 - p̂)/n]

With p̂ = 0.52 and z* ≈ 1.96, the margin of error (ME) is:

ME = 1.96  √[0.52  0.48 / 1000] ≈ 1.96  √(0.2496 / 1000) ≈ 1.96  0.0158 ≈ 0.031

Thus, the 95% confidence interval is:

(0.52 - 0.031, 0.52 + 0.031) ≈ (0.489, 0.551)

The lower bound surpasses 0.5, suggesting that more than half favor the Republican candidate within this interval, with high statistical confidence.

To estimate the probability that the Democratic candidate is actually leading, given the observed data, we evaluate the likelihood that the true proportion exceeds 0.5. Since the interval includes 0.5, but the point estimate is slightly above, the probability that the Democratic support surpasses that of Republicans can be approximately 50%, recognizing the margin of sampling error.

Repeating the analysis with a larger sample of 3,000 voters and the same proportions reduces the margin of error, providing more precise estimates. The increased sample reduces the standard error, narrowing the confidence interval and increasing certainty about the political support levels.

Analysis of Customer Data from Lincolnville School District

Constructing confidence intervals for the mean bus maintenance costs and odometer miles involves similar principles: collecting sample means and their standard errors, then applying the formula for the confidence interval. Given the data, suppose the sample mean maintenance cost is calculated, and the sample size is n=30 buses. The confidence interval at 95% confidence is:

CI = x̄ ± 1.96 * (s / √n)

Similarly, for odometer miles, the same process applies. These intervals provide estimates for the respective population means, enabling informed decisions regarding maintenance planning and fleet usage.

Case Study: Debit Card Usage at Century National Bank

To determine the proportion of customers using debit cards, a sample is taken (e.g., 60 customers), and the proportion who use these cards is calculated. Using this sample proportion, a 95% confidence interval is constructed with:

CI = p̂ ± z  √[p̂(1 - p̂)/n]

If the interval's lower bound exceeds 0.5, it suggests that more than half of the bank's customers use debit cards. Based on sample data, the bank possibly has a majority of customers utilizing this service, which has implications for marketing, security, and operational strategies.

Overall, applying statistical inference to these real-world data sets allows for meaningful conclusions, informed decision-making, and effective resource allocation across diverse sectors such as retail pricing, public transportation, healthcare, politics, and banking services.

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