Survey Of American Planning Long Summer Vacations In 2002

A Survey Of American Planning Long Summer Vacations In 2002 Reveale

A survey of American planning long summer vacations in 2002 revealed a mean planned expenditure of $1076. Assume that this mean is based on a random sample of 300 Americans who were planning long summer vacations in 2002 and that the sample standard deviation was $345. What is the point estimate of the mean planned expenditure by all Americans taking long summer vacations in 2002? What is the 95% margin of error for the above mean planned expenditure? Construct a 99% confidence interval for the mean planned expenditure of 2002. What does the 99% confidence interval mean in this context? What is the maximum error (margin of error) estimate for part c)? Suppose the standard deviation of all Americans’ planned expenditure of 2002 is believed to be $400 according to some experts. What, then, is the 99% confidence interval for the mean planned expenditure of 2002? What is the maximum error (margin of error) of estimate for part f)?

It is claimed that a new treatment is more effective than the standard treatment for prolonging the lives of terminal cancer patients. The standard treatment has been in use for a long time, and from records in medical journals, the mean survival period is known to be 4.2 years. The new treatment is administered to 80 patients and their duration of survival recorded. The sample mean and the standard deviation are found to be 4.5 and 1.1 years, respectively. Answer the following questions: Define the context (population parameter) in this problem. Formulate the null and alternative hypotheses. Is the alternative hypothesis one-sided or two-sided? State the Type I error and Type II error for this problem and explain why Type I error is more serious. State the Z-test statistic for this problem and identify its distribution. What value of the Z-test (small, large, or both) will lead to rejecting the null hypothesis? Suppose the significance level is 0.05. Determine the critical value for the rejection region and write down the decision rule. Calculate the Z-test statistic using the data provided. Decide whether or not to reject the null hypothesis based on this value, and explain why. Calculate the P-value and perform the hypothesis test based on it. Finally, interpret the conclusion in plain English in the context of the problem.

The steel factory produces iron rods that are supposed to be 36 inches long. The machine does not produce each rod exactly 36 inches, but the mean length when functioning properly is 36 inches with a standard deviation of 0.05 inches. A sample of 40 rods is tested to determine whether the process is in control. The null hypothesis is that the mean length is 36 inches, against the alternative that it is not. The sample mean length is 36.015 inches. Analyze this data: Write hypotheses, determine if the alternative is one-sided or two-sided, state the errors, compute the Z-statistic, identify the rejection criterion at 1% significance level, and decide whether to reject the null hypothesis. Interpret the results in simple English.

Paper For Above instruction

The 2002 survey on American planning long summer vacations provided insights into consumers’ spending habits, revealing a mean expenditure of $1076 based on a sample of 300 individuals with a sample standard deviation of $345. The point estimate for the population mean expenditure is therefore $1076, serving as the best estimate based on the sample data. To assess the variability and reliability of this estimate, constructing confidence intervals is essential. The 95% margin of error can be computed using the formula for the margin of error (ME) for a population mean when the population standard deviation is unknown and the sample size is large, which involves the t-distribution or the normal distribution approximation; however, given the large sample size, the standard normal distribution can be applied here. The margin of error at a 95% confidence level is approximately $392.90, leading to a confidence interval ranging from approximately $683.10 to $1472.90. This interval indicates that we are 95% confident that the true mean expenditure lies within this range.

Furthermore, for a 99% confidence interval, the margin of error increases due to the larger z-value corresponding to the higher confidence level. If the standard deviation of all American expenditures is believed to be $400, then the 99% confidence interval would extend from approximately $510.16 to $1641.84, with a margin of error of about $574.84. These intervals demonstrate how higher confidence levels widen the range, reflecting greater uncertainty.

In the context of health research, a new treatment for terminal cancer aims to improve survival rates. The known average survival time with the standard treatment is 4.2 years, and a sample of 80 patients undergoing the new treatment has a mean survival of 4.5 years with a standard deviation of 1.1 years. The parameter of interest, the mean survival time difference, is evaluated through hypothesis testing. The null hypothesis posits no difference or that the mean survival is 4.2 years, while the alternative suggests that the new treatment extends survival beyond this duration. Since the claim is that the new treatment is more effective, this is a one-sided test, testing whether the mean survival exceeds 4.2 years. The risks of Type I error (false positive) and Type II error (false negative) are considered; Type I error is more serious here because wrongly concluding the treatment is effective could lead to widespread adoption of a potentially unproven approach with significant health implications.

Calculating the Z-statistic involves subtracting the hypothesized mean from the sample mean and dividing by the standard error, which accounts for the sample size and standard deviation. The resulting Z-value exceeds the critical value at the 5% significance level, leading us to reject the null hypothesis. The P-value, obtained from the standard normal distribution, supports this conclusion, suggesting that the new treatment significantly prolongs survival compared to the standard.

Regarding the manufacturing process, the quality control measure involves testing the mean length of rods produced by a machine reputed to produce 36-inch rods with a known standard deviation of 0.05 inches. A sample of 40 rods yields a mean of 36.015 inches. The hypotheses test whether the process is in control, with the null asserting the mean length is 36 inches and the alternative that it is not. Since the concern is whether the process remains accurate or not, the test is two-sided. At a 1% significance level, the critical Z-value is approximately ±2.576. Calculations of the Z-statistic from the sample data demonstrate whether the process deviates significantly from the target length. The result indicates the null hypothesis should not be rejected, implying the manufacturing process remains within acceptable limits. The interpretation emphasizes that, statistically, the process is under control, and no adjustments are currently necessary.

Together, these three statistical analyses exemplify how hypothesis testing and confidence interval construction can provide actionable insights across different fields—economic planning, medical research, and manufacturing quality control. Each scenario underscores the importance of understanding the underlying parameters, the structure of hypotheses, and interpreting the statistical results in a practical context to support sound decision-making.

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