Performance Lawn Equipment: What Proportion Of Customers Rat

Performance Lawn Equipmentwhat Proportion Of Customers Rate The Compan

Performance Lawn Equipmentwhat Proportion Of Customers Rate The Compan

Performance Lawn Equipment What proportion of customers rate the company with “top box” survey responses (which is defined as scale levels 4 and 5) on quality, ease of use, price, and service in the 2012 Customer Survey worksheet? How do these proportions differ by geographic region? What would be a confidence interval for an additional sample of mower test performance in the worksheet Mower Test? For the data in the worksheet Blade Weight, what is the sampling distribution of the mean, the overall mean, and the standard error of the mean? Is a normal distribution an appropriate assumption for the sampling distribution of the mean? How many blade weights must be measured to find a 95% confidence interval for the mean blade weight with a sampling error of at most 0.2? What if the sampling error is specified as 0.1?

Paper For Above instruction

The evaluation of customer satisfaction and product performance for Performance Lawn Equipment involves multiple statistical analyses, notably the estimation of proportionate customer ratings, comparison across regions, and inferential statistics for product measurements such as blade weight. This paper delves into each component, providing statistical reasoning and calculations based on the underlying survey and testing data, assuming such data is representative and appropriately sampled.

Proportion of Customers Giving “Top Box” Ratings

Based on the 2012 Customer Survey worksheet, the proportion of customers assigning the highest ratings ("top box", levels 4 and 5) to the company's quality, ease of use, price, and service can be estimated by dividing the number of respondents giving these ratings by the total number of respondents for each category. For example, if in the quality category, 800 out of 1000 respondents rated 4 or 5, then the proportion is 0.8 (80%). Similar computations are conducted across the other categories and regions.

To compare these proportions across geographic regions, stratified analysis or chi-square tests for independence may be used to determine if significant differences exist. Suppose the survey divides responses into regions such as North, South, East, and West; calculating separate proportions for each allows for comparison. For instance, if Region North reports a 0.85 proportion, and Region South reports 0.75, a hypothesis test can be performed:

H0: p_North = p_South

Ha: p_North ≠ p_South


using a two-proportion z-test, where the test statistic is computed based on the pooled proportion and the difference in sample proportions. This analysis reveals whether regional differences are statistically significant.

Confidence Interval for Additional Mower Test Performance Sample

Assuming the sample data from the worksheet Mower Test provides the mean test performance score and standard deviation, a confidence interval can be constructed for future samples. For example, with a sample mean of μ̂ and a standard deviation s, the confidence interval at the 95% level is:

CI = μ̂ ± t*(s/√n)

where t* is the critical value from the t-distribution with n-1 degrees of freedom, based on the sample size n. This interval estimates the range within which future mower performance scores are expected to fall with 95% confidence.

Sampling Distribution of Blade Weight

The data in the worksheet Blade Weight can be analyzed to determine the sampling distribution of the mean blade weight. When multiple measurements are taken across blades, the sample mean serves as an estimator of the population mean. The standard error of the mean (SEM) is calculated as:

SEM = s / √n

where s is the sample standard deviation, and n is the number of measurements. The overall mean from all measurements provides an estimate of the central tendency, while the SEM indicates the variability of the sample mean estimate.

The Central Limit Theorem supports assuming a normal distribution for the sampling distribution of the mean, especially when n is large (typically n ≥ 30). If the blade weights are approximately normally distributed or the sample size is sufficiently large, then normality is a reasonable assumption, facilitating the use of normal-based inference.

Sample Size for Desired Confidence Interval Precision

To determine how many blade weights need to be measured for a specified margin of error (sampling error), the formula for sample size n is used:

n = (z* s / E)^2

where z* is the z-value corresponding to the confidence level (1.96 for 95%), s is an estimate of the population standard deviation, and E is the desired margin of error.

For a margin of error E = 0.2, and assuming an estimated standard deviation s = 1.0 (hypothetically),

n = (1.96 * 1 / 0.2)^2 ≈ 96

Similarly, for E = 0.1,

n = (1.96 * 1 / 0.1)^2 ≈ 385

These results suggest measuring approximately 96 blades to achieve a margin of 0.2, or about 385 blades for a margin of 0.1, ensuring the precision of the mean blade weight estimate.

Conclusion

Effective analysis of customer ratings and product measurements requires combining descriptive statistics, hypothesis testing, and confidence interval estimation. Understanding regional differences in customer perceptions can guide targeted improvements, while precise measurement of blade weight supports quality control. Applying appropriate statistical models, such as the normal distribution assumptions for large samples, ensures reliable inference and decision-making.

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