The lifetimes (in miles) of a certain brand of automobile tires follow a normal distribution with a mean (µ) of 40,000 miles and a standard deviation (σ) of 2,000 miles. The manufacturer plans a guarantee policy where they will replace any tire that lasts less than a certain number of miles, aiming to cover only 1% of all tires sold. The task is to determine this cutoff mileage, which corresponds to the shortest 1% of tire lifetimes. This involves identifying the 1st percentile in the normal distribution with given parameters.
Additionally, the content includes various statistical concepts and questions such as the estimation of regression models, hypothesis testing, confidence intervals, and interpretations of statistical outputs. For example, one statement suggests that the model Y = β0 + β1X + β2X² cannot be estimated using Excel due to the nonlinear term, which is false. Also, the critical value for a right-tailed test with n = 25 at α = 0.05 is approximately 1.711, not 1.645 or 0.0179, which are incorrect choices in this context.
In a hypothesis test concerning the weight of Sonora Bars, the null hypothesis asserts that the mean weight equals 56 grams, and the alternative hypothesis suggests it may be less—indicating a test for a decrease in mean weight. The sample data includes a mean of 55.82 grams with known process standard deviation of 0.77 grams and a sample size of 49. The appropriate hypotheses are H0: µ = 56 and H1: µ < 56.
Furthermore, a confidence interval for the average annual expenses on books and class materials is calculated from a sample of 36 students, with a sample mean of $850 and a standard deviation of $54. The 99% confidence interval is approximately from $826.82 to $873.18, providing a range within which the true population mean likely falls.
Lastly, the statement regarding SSE (Sum of Squares for Error) in regression indicates that if SSE is near zero, the model explains nearly all variability in the data, implying a good fit; thus, the statement claiming it indicates a poor fit is false.
Paper For Above instruction
The problem of determining the cutoff mileage for the tire replacement guarantee exemplifies the application of basic properties of the normal distribution. The manufacturer's concern is to set a threshold such that only the lowest 1% of tire lifetimes fall below this point, ensuring that the financial risk remains minimal while maintaining customer satisfaction. Given the normal distribution with µ = 40,000 miles and σ = 2,000 miles, the relevant percentile can be found using standard normal distribution tables or statistical
software.
To find the cutoff, we determine the z-score corresponding to the 1st percentile. From standard normal distribution tables, the z-score for the 1% lower tail is approximately -2.33. Using the z-score formula:
z = (X - µ) / σ
we solve for X:
X = µ + z * σ = 40,000 + (-2.33) * 2,000 = 40,000 - 4,660 = 35,340 miles.
Thus, the shortest 1% of tire lifetimes lasts less than approximately 35,340 miles. The manufacturer should set the guarantee threshold at about 35,340 miles, replacing any tires that fail before this mileage. This ensures coverage of approximately 1% of the worst-performing tires, aligning with their policy goals.
In statistical modeling, the question about the regression model Y = β0 + β1X + β2X² emphasizes the limitations of Excel in estimating non-linear models. Since the presence of a quadratic term introduces non-linearity, standard linear regression procedures like those in basic Excel functions cannot directly estimate such models without specialized add-ins or transformations, making the statement false. Advanced statistical software such as R or SAS is necessary for such models.
The critical value for a right-tailed t-test with a sample size of 25 (degrees of freedom = 24) at a significance level of α = 0.05 can be found in t-distribution tables or statistical software. Approximately, the t-critical value is 1.711, aligning with the options provided, whereas 1.645 corresponds to the z-distribution for large sample sizes, and 0.0179 has no relevance here.
The hypothesis testing involving the weight of Sonora Bars involves testing whether the true mean weight is less than the specified 56 grams. Since the producer suspects that the mean might be smaller, the appropriate hypotheses are:
Null hypothesis, H0: µ = 56 grams
Alternative hypothesis, H1: µ < 56 grams
This setup allows for testing whether the process is producing lighter bars than intended, which could signal issues in quality control.
Regarding confidence intervals, the calculation from the sample data yields an interval estimating the population mean of students’ annual expenses on books and class materials. With n = 36, mean = $850,
and standard deviation = $54, and using the z-distribution for 99% confidence, the interval is computed as:
CI = mean ± Z * (σ/√n) = 850 ± 2.576 * (54/6) ≈ 850 ± 2.576 * 9 = 850 ± 23.184 which results in the interval approximately from $826.82 to $873.18. This range indicates where the true mean likely resides with 99% confidence.
Finally, the statement on SSE suggests that if the sum of squared errors in a regression is near zero, the model accurately captures the variability. This would normally indicate a good fit, not a poor one. Thus, the statement claiming the opposite is false, as near-zero SSE reflects a highly effective model fit.
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