Sample Of 81 Observations Taken From A Normal Population

Sample Of 81 Observations Is Taken From A Normal Population With

Determine the 95 percent confidence interval for the population mean when a sample of 81 observations is taken from a normal population with a standard deviation of 5, and the sample mean is 40.

Construct a 95 percent confidence interval for the mean number of misspelled words per essay based on Dr. Patton’s data, where the population standard deviation is 2.44 words, the sample size is 40 students, and the sample mean is 6.05 words.

Use Appendix B.2 to find the critical t-values for the following conditions: (a) a sample size of 15 with a 95% confidence level; (b) a sample size of 24 with a 98% confidence level; (c) a sample size of 12 with a 90% confidence level.

The U.S. Dairy industry wants to estimate the mean yearly milk consumption. A sample of 16 people has a mean of 60 gallons and a standard deviation of 20 gallons. Determine: (a) the best estimate of the population mean; (b) why the t-distribution is needed; (c) the t-value for a 90% confidence interval; (d) the 90% confidence interval; (e) whether the population mean could reasonably be 63 gallons.

Ms. Maria Wilson is considering running for mayor. A survey of 400 voters shows 300 support her. Estimate the population proportion, construct a 99% confidence interval for this proportion, and interpret the findings.

To estimate the population mean within 5 units at a 99% confidence level, given the population standard deviation is 15, determine the required sample size.

To estimate the population proportion within ±0.10 with a 99% confidence level, given the proportion estimate of 0.45, determine the necessary sample size.

Paper For Above instruction

Statistical inference plays a crucial role in understanding populations based on sample data. This paper explores several applied scenarios involving confidence intervals, hypothesis testing, and sample size calculations using normal and t-distributions. Each scenario underscores different aspects of these statistical concepts and their practical significance in fields such as education, industry, and politics.

Confidence Intervals for the Population Mean

Confidence intervals provide a range within which the true population parameter is expected to fall with a specified level of confidence. For instance, when estimating the mean from a sample of 81 observations from a normal population with a known standard deviation of 5 and a sample mean of 40, the 95% confidence interval can be calculated using the z-distribution, since the population standard deviation is known. The formula is:

CI = x̄ ± z*(σ/√n)

where x̄ is the sample mean, σ is the population standard deviation, n is the sample size, and z is the critical value from the standard normal distribution for the desired confidence level. Here, z for 95% confidence is approximately 1.96, leading to:

CI = 40 ± 1.96(5/√81) = 40 ± 1.96(5/9) = 40 ± 1.96*0.555 ≈ 40 ± 1.089

Thus, the interval is approximately (38.911, 41.089).

Similarly, Dr. Patton's analysis of misspelled words involves constructing a confidence interval for the population mean number of misspelled words, using the known standard deviation of 2.44, a sample size of 40, and a sample mean of 6.05. The z-value remains the same at 1.96, and the interval is:

6.05 ± 1.96(2.44/√40) ≈ 6.05 ± 1.96(2.44/6.3246) ≈ 6.05 ± 1.96*0.386 ≈ 6.05 ± 0.756

which results in an interval of approximately (5.294, 6.806).

Using the T-Distribution and Critical Values

The t-distribution comes into play when the population standard deviation is unknown, or the sample size is small. Appendix B.2 provides critical t-values for different sample sizes and confidence levels, essential for accurate interval estimation. For example, with a sample size of 15 at 95% confidence, the t-value can be found in the t-table under 14 degrees of freedom. For a 98% confidence level with n=24 (23 df), and for n=12 (11 df) at 90%, the t-values are higher than the corresponding z-values due to added uncertainty with smaller samples.

Estimating Population Parameters in Industry

In the case of milk consumption in the U.S. Dairy industry, a samplemean of 60 gallons with a standard deviation of 20 gallons from 16 individuals provides a basis for estimation. The best estimate of the population mean is simply the sample mean, 60 gallons. To determine the confidence interval, the t-distribution is used because the population standard deviation is unknown, and the sample size is small. The t-value for 90% confidence with 15 degrees of freedom (n-1) is approximately 1.753. The interval is:

60 ± 1.753(20/√16) = 60 ± 1.7535 = 60 ± 8.765

, resulting in (51.235, 68.765). The interval indicates uncertainty and suggests the true mean could be within this range, affecting industry planning and policy-making. The assumption here is that the sample is representative and that the data follows a normal distribution, which justifies the use of the t-distribution.

Estimating Population Proportions

For the political support survey, 300 out of 400 voters favor Ms. Wilson, giving a sample proportion of 0.75. The confidence interval for the proportion uses the formula:

CI = p̂ ± z*√(p̂(1 - p̂)/n)

where p̂ is the sample proportion. For a 99% confidence level, z* is approximately 2.576. Calculating:

0.75 ± 2.576√(0.750.25/400) = 0.75 ± 2.576√(0.1875/400) = 0.75 ± 2.576√0.00046875 ≈ 0.75 ± 2.576*0.02166 ≈ 0.75 ± 0.0558

The interval ranges roughly from 0.6942 to 0.8058, indicating with high confidence that support among the entire voter population lies within this range. This supports strategic political decisions based on voter preferences.

Sample Size Calculations

To estimate the population mean within 5 units at 99% confidence level with a standard deviation of 15, the required sample size n is found using:

n = (z*σ/E)^2

where E is the desired margin of error (5), and z* for 99% confidence is approximately 2.576. Plugging in the numbers:

n = (2.57615/5)^2 = (2.5763)^2 = (7.728)^2 ≈ 59.75

So, at least 60 observations are necessary.

Similarly, to estimate the proportion within ±0.10 with 99% confidence when the estimated proportion is 0.45, the sample size n is computed using:

n = (z/E)^2 p̂(1 - p̂)

which leads to:

n = (2.576/0.10)^2 0.450.55 = (25.76)^2 0.2475 ≈ 664.4 0.2475 ≈ 164.55

Therefore, approximately 165 observations are needed to achieve this accuracy in estimating the proportion.

Conclusion

In summary, the application of confidence intervals, t-distribution, and sample size estimation forms the backbone of effective statistical analysis in diverse fields. Whether estimating averages or proportions, understanding underlying assumptions, like normality and independence, is essential for valid inferences. These methods enable researchers and policymakers to make informed decisions grounded in statistical evidence, emphasizing the importance of correct application and interpretation of these statistical tools.

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