Problem 1510a: What Is The Standard Deviation Of Juan's Mean

Problem 1510a What Is The Standard Deviation Of Juans Mean Result

Problem 15.10 : a) What is the standard deviation of Juan’s mean result: We know that the standard deviation of mean related to standard deviation of distribution by σmean = standard deviation/ square root (n) ……..(1) Given, n=4 and standard deviation = 10, So we have σmean =10/sqrt(4) = 10/2 = 5 b) Standard deviation of sample mean = 2, find n From (1) we have 2=10/sqrt(n) Sqrt(n) = 10/2 = 5 n =25 When there is a single measurement, it may not be close to the population mean. When averaging the measurements, it gives a result which will be nearer to the population mean. Problem 15.12: a) n=50. Since the sample size is greater than 30 we may use central limit theorem. We are given, mean = 0.5 and standard deviation = 0.7 We have mean =0.5 Also, we have from (1) we have σmean = standard deviation/ square root (n) = 0.7/sqrt(50) = 0.9899 b) Prob( average no of moths in 50 traps is greater than 0.6) Sample mean = 0.6 population mean = 0.5 standard deviation = 0.7 We have Z = (sample mean – population mean)/(σ/sqrt(n) Z = (0.6 – 0.5)/(0.7/sqrt(50)) =1.01 P(Z>1.01) = 0.1562 Problem 15.28: We are given that the scores follow normal distribution. We have mean µ=25 and standard deviation = 6.5 a) Probability (score of a student is between 20 and 30) We now standardize the scores. P((20-25/(6.5/sqrt(1))2 = 0.5588 = 0.56 . b) n=25….sampling distribution of x bar sampling distribution of x bar = sigma / sqrt(n) = 6.5/sqrt(25) = 1.3 c) If n= 25, we need to find P((20-25/(6.5/sqrt(25))Z 10.0833 ± 1.645(0.1/sqrt(6) = (10.01586,10.15075) is the confidence interval for true conductivity. 17.6) The average income of American women who work full-time and have only a high school degree is $31,666. You wonder whether the mean income of female graduates from your local high school who work full-time but have only a high school degree is different from the national average. You obtain income information from an SRS of 62 females graduates who work full-time and have only a high school degree and find that x=$30,052. What are your null and alternative hypothesis? ANSWER: Null hypothesis: Average income of American women who work full – time and only have a high school degree is $31,666. That is µ=31,666 Alternative hypothesis: Average income of American women who work full – time and only have a high school degree is different from 31,666. That is µâ‰ $31,666 Problem 17.3: a) Null hypothesis: Students on an average study for 15 hours a week. i.e., µ=15 hours Alternative hypothesis: Students on an average study for more than 15 hours a week i.e.,µ>15 hours Problem 18.4 : n=477 Sample mean = 157 Standard deviation = 35 Z for 95% confidence interval is 1.96. The confidence interval is (Sample mean ± Z(α/2)Standard error) Standard error = standard deviation/sqrt(n) =35/sqrt(n) =35/√477 ≈ 1.60. Confidence interval = (157±(1.96)(1.60)) =(153.864,160.136). Since we only have the sample mean and standard deviation, we cannot definitively confirm the population mean unless it falls within this interval. Problem 16.28: We understand that the Standard error = standard deviation/ sqrt(n). When sample size increases, the standard error decreases, and vice versa. The margin of error is Zstandard deviation/sqrt(n). For example, with Z=1.96 at 95% confidence level: for n=8, margin of error ≈ 0.693σ; for n=50, margin ≈ 0.277σ. Larger samples reduce the margin of error, increasing confidence in estimates. Problem 18.14: a) Power of the test is the probability of correctly rejecting the null hypothesis when it is false. In this case, TUDA measures reading progress as 243; a sample yields 255, prompting a test. The probability of correctly rejecting the null (power) is 0.29, which is low. b) A power of 0.8 is generally desired for reliable results. Since this test has a power of 0.29, it indicates low reliability, and larger sample sizes are recommended for more accurate testing of whether reading progress exceeds 243. The account balances of Viaduct Co. as of December 31, 20xx, include various assets and liabilities such as cash, receivables, inventory, and expenses; the data needed for year-end adjustments include inventory counts, expired insurance, and accrued expenses. The task involves preparing a worksheet, journalizing adjustments, and closing entries for the fiscal year ending December 31, 2006.

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The calculation of the standard deviation of Juan's mean result illustrates fundamental principles of statistical inference, particularly how sample size influences the variability of mean estimates. When dealing with sample means, the standard deviation of the mean, also known as the standard error (σₘ), becomes crucial because it quantifies the expected spread of sample means around the true population mean. The mathematical relationship expresses this as σₘ = σ / √n, where σ is the population standard deviation, and n is the sample size. In the scenario provided, with a standard deviation (σ) of 10 and a sample size (n) of 4, the standard error computes to 5, indicating that the mean of such samples would, on average, deviate from the population mean by this amount.

Further, based on the second part of the problem, if the standard deviation of the sample mean is specified as 2 and the original population standard deviation is known to be 10, one can determine the sample size by rearranging the formula. Setting σₘ = 2 yields n = (σ / σₘ)² = (10 / 2)² = 25. This highlights how increasing the sample size reduces the variability of the mean, thereby providing a more precise estimate of the population parameter.

Extending this understanding to large samples, the central limit theorem justifies the use of normal distribution approximations even when the underlying data are not normally distributed, provided the sample size exceeds 30. For instance, in a problem involving the mean number of moths in traps with n=50, a known population mean of 0.5, and a standard deviation of 0.7, the standard error reduces to approximately 0.9899, allowing for inference about the population mean. Calculating the probability that the sample mean exceeds a certain threshold (e.g., 0.6) involves standardizing and referring to the standard normal distribution, resulting in a probability of approximately 15.62%, indicating moderate likelihood of observing such a deviation.

The application of normal distribution assumptions extends further in analyzing test scores, such as a student scoring between 20 and 30 on a test with a mean of 25 and a standard deviation of 6.5. Standardizing these bounds yields Z-scores approximately ±0.77. Using normal distribution tables, the probability that a score falls within this interval is about 56%. When considering sample sizes, for example n=25, the sampling distribution of the mean has a standard error of 1.3, and the probability of scores between 20 and 30 becomes nearly 99.99%, reflecting the increased precision of the mean estimate with larger samples.

Confidence intervals offer another vital application of standard errors. For example, calculating a 90% confidence interval for the true conductivity based on a sample mean of 10.0833 and a standard deviation of 0.1 in six measurements yields an interval of approximately (10.01586, 10.15075). Such intervals provide a range within which the true parameter is likely to fall, with the confidence level indicating the proportion of such intervals that would contain the parameter over repeated samples.

In hypothesis testing, defining null and alternative hypotheses forms the foundation. For instance, considering the average income of women with different education levels, the null hypothesis posits equality to the national average, while the alternative suggests a difference, often formulated as µ ≠ 31,666. Testing this hypothesis requires computing a test statistic, such as a z-score, and comparing it to critical values, which dictate whether to reject the null hypothesis. Similar frameworks apply to other contexts like students’ study hours or classroom test scores, where hypotheses are framed to assess whether means exceed or differ from specified benchmarks.

The influence of sample size on the margin of error is fundamental in survey estimation. The margin of error at a given confidence level is proportional to Z * (σ / √n), emphasizing that increasing n reduces the margin, thus increasing the precision of the estimate. For example, with a Z-value of 1.96, increasing sample size from 8 to 50 significantly decreases the margin of error, illustrating the inverse relationship between sample size and estimation uncertainty. Statisticians favor larger samples because they produce more reliable estimates with narrower confidence intervals.

Finally, the concept of statistical power—the probability of correctly rejecting a false null hypothesis—is crucial in experimental design. Given a measured reading progress of 255, when the null hypothesis states 243, the probability of correctly detecting a real difference is the power. A low power (e.g., 0.29) suggests a high likelihood of Type II errors, meaning real effects may go unnoticed. To achieve better detection capabilities, increasing the sample size or improving measurement precision is essential, reinforcing the importance of adequate sample size in statistical testing.

The comprehensive analysis of the Viaduct Co. accounts and adjustments emphasizes diligent bookkeeping and understanding of financial principles. Preparing a worksheet involves listing all accounts systematically, ensuring that debit and credit entries balance after the application of year-end adjustments. Journalizing adjustments for inventory, expired insurance, accrued expenses, and depreciation ensures that financial statements accurately reflect the company's financial position. Subsequently, closing entries are recorded to transfer temporary account balances to retained earnings, establishing a clean slate for the next accounting period. These procedures are foundational to accurate financial reporting, underscoring the importance of meticulous accounting practices in maintaining organizational transparency and accountability.

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