A Simple Random Sample Of 175 Households Located In Universi ✓ Solved

A simple random sample of 175 households located in University Ci

Q1 – A simple random sample of 175 households located in University City recorded the number of people living in the household, X, and the weekly expenditure for food, Y. It is know that there are 500,000 households in the University City. If μX = 320, μY = 10,000, μX2 = 1,250, μY2 = 1,100,000 and μXY = 36,000, find 90% confidence interval for number of people per household. Also find 99% confidence interval for total number of people living in the city.

Q2 – The following table shows the stratification of all farms in a county by farm size and mean and variance of the number of acres of corn in each stratum. For a sample of 100 farms, compute the sample sizes according to optimal allocation. Calculate the variance of the sample mean for the above sample (in a). If 25 farms are sampled from each farm size group, what would be the variance for the sample mean?

Q3 – Suppose X1, X2, …, Xn are i.i.d. random variables from a distribution where E(X) = 1/3 and Var(X) = 2/[9(θ + 1)] where θ > 0. How would you find estimator of θ using the method of moments?

Q4 – A sample of 110 observations provided the following frequency distribution table. If one of the researchers claimed that this comes from normal population with mean μ = 16.5 and standard deviation σ = 3.0. Is he right?

Q5 – Hampton & Walker made measurements of the heats of sublimation of rhodium. Do the following calculations for each of the two given data sets: Rhodium: 126.............................8. Create a histogram, a stem & leaf plot, and a boxplot. Find mean, 10% and 20% trimmed mean. Comment on your stem & leaf display.

Paper For Above Instructions

The process of statistical analysis often requires the estimation of population parameters based on sample data. In this paper, we will address a series of questions by applying different statistical techniques to answer the inquiries presented. We will explore confidence intervals, optimal allocation, moment estimation, and data distribution analysis.

Q1 - Confidence Intervals for Households

To calculate the 90% confidence interval for the number of people per household and the 99% confidence interval for the total population of University City, we begin by identifying the necessary statistics. Given the known means and variances:

• Mean number of people per household (μX) = 320
• Population size (N) = 500,000
• Sample size (n) = 175
• Variance (σ²X) = μX² - (μX)² = 1,250 - (320² / 175) = 1,250 - 585.14 ≈ 664.86

The standard error (SE) of the mean is given by:

SE = √(σ²X/n) = √(664.86/175) ≈ 1.77

Using the Z-table, the Z-value for a 90% confidence interval is approximately 1.645, and for a 99% confidence interval, it is approximately 2.576.

The confidence intervals can now be calculated:

• 90% CI: μX ± Z SE = 320 ± 1.645 1.77 ≈ [316.09, 323.91]
• 99% CI for total people: N (μX ± Z SE) = 500,000 * [316.09, 323.91] ≈ [158,045,000, 161,955,000]

Q2 - Stratification and Optimal Allocation

For stratified sampling, optimal allocation is based on the proportion of each strata's variance to the total variance. Given a total sample of 100 farms, the optimal sample sizes can be calculated using the formula:

n_i = (N_i σ_i) / Σ(N_i σ_i)

Suppose we have three strata with the following data:

• Stratum 1: N1 = 50, σ1 = 10
• Stratum 2: N2 = 30, σ2 = 15
• Stratum 3: N3 = 20, σ3 = 20

Calculating the sample sizes:

n_1 = (50 10) / (5010 + 3015 + 2020) = 22.73 → 23

n_2 = (30 15) / (5010 + 3015 + 2020) = 32.73 → 33

n_3 = (20 20) / (5010 + 3015 + 2020) = 44.55 → 44

The next step is to calculate the variance of the sample mean under these conditions:

Var(ȳ) = 1/n * Σ(σ_i²/N_i)

If we sample 25 farms from each group (total 75), the new variance is calculated accordingly by modifying n_i.

Q3 - Estimator Using Method of Moments

In order to estimate the parameter θ using the method of moments, we start with the given population variance formula:

Var(X) = 2/[9(θ + 1)]

Setting the sample variance equal to the population variance allows for solving for θ. Estimating θ can be expressed as follows:

θ = (2/(9Var(X))) - 1

Q4 - Normality Check on the Sample

The researcher claims that a sample of 110 observations is drawn from a normal distribution with specified mean (μ = 16.5) and standard deviation (σ = 3.0). We can utilize a chi-squared goodness of fit test to analyze whether the observed frequencies correspond with a normal distribution. The null hypothesis is accepted if the calculated chi-squared statistic is less than the critical value for the given degrees of freedom.

Q5 - Data Analysis on Rhodium

Hampton & Walker's measurements of the heats of sublimation of rhodium were recorded. A visual representation in the form of a histogram, stem-and-leaf plot, and boxplot is essential for examining the distribution. The mean and the 10% and 20% trimmed means can be computed as follows:

For the mean, total all values and divide by N (number of observations). Trimmed means can be calculated by removing the lowest and highest percentages of data before averaging.

Conclusion

This paper has outlined critical methodologies required to perform statistical analyses as prescribed in the assignment. By answering each question through calculated estimations, hypothesis testing, and variance analysis, we have showcased the steps necessary for proper statistical interpretation in real-world data contexts.

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