Question

1. 2/11 pointsthirty small communities in connecticut (population near 10,000 each) gave an average of \\( \bar{x}=137.4 \\) reported cases of larceny per year. assume that \\( \sigma \\) is known to be 40.7 cases per year.(a) find a \\( 90 \\% \\) confidence interval for the population mean annual number of reported larceny cases in such communities. what is the margin of error? (round your answers to one decimal place.)lower limitupper limitmargin of error(b) find a \\( 95 \\% \\) confidence interval for the population mean annual number of reported larceny cases in such communities. what is the margin of error? (round your answers to one decimal place.)lower limitupper limitmargin of error(c) find a \\( 99 \\% \\) confidence interval for the population mean annual number of reported larceny cases in such communities. what is the margin of error? (round your answers to one decimal place.)lower limitupper limitmargin of error(d) compare the margins of error for parts (a) through (c). as the confidence levels increase, do the margins of error increase?\\( \bigcirc \\) as the confidence level increases, the margin of error increases.\\( \bigcirc \\) as the confidence level increases, the margin of error remains the same.\\( \bigcirc \\) as the confidence level increases, the margin of error decreases.

Explanation:

Step1: Calculate margin of error for 90% confidence interval

The formula for margin of error \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\). For 90% confidence level, \(\alpha=1 - 0.90=0.10\), \(\alpha/2 = 0.05\), and \(z_{\alpha/2}=z_{0.05}\approx1.645\) (from standard normal table). Given \(\sigma = 40.7\), assume \(n = 30\) (since it's about 30 communities). Then \(E=1.645\times\frac{40.7}{\sqrt{30}}\approx1.645\times7.43\approx12.2\) (rounded to one decimal place).
Lower limit \(=\bar{x}-E = 137.4-12.2 = 125.2\) (incorrect in the given wrong answer, correct calculation). Upper limit \(=\bar{x}+E=137.4 + 12.2=149.6\) (incorrect in given wrong answer). Margin of error \(E\approx12.2\) (but given wrong as 5.5).

Step2: Calculate for 95% confidence interval

For 95% confidence level, \(\alpha = 1- 0.95=0.05\), \(\alpha/2=0.025\), \(z_{\alpha/2}=z_{0.025}=1.96\). \(E = 1.96\times\frac{40.7}{\sqrt{30}}\approx1.96\times7.43\approx14.6\). Lower limit \(=137.4-14.6 = 122.8\) (incorrect in given wrong answer). Upper limit \(=137.4 + 14.6=152.0\) (incorrect in given wrong answer). Margin of error \(E\approx14.6\) (but given wrong as 7).

Step3: Calculate for 99% confidence interval

For 99% confidence level, \(\alpha=1 - 0.99=0.01\), \(\alpha/2=0.005\), \(z_{\alpha/2}=z_{0.005}=2.576\). \(E=2.576\times\frac{40.7}{\sqrt{30}}\approx2.576\times7.43\approx19.1\). Lower limit \(=137.4-19.1=118.3\) (incorrect in given wrong answer). Upper limit \(=137.4 + 19.1 = 156.5\) (incorrect in given wrong answer). Margin of error \(E\approx19.1\) (but given wrong as 9).

Step4: Analyze margin of error vs confidence level

As the confidence level increases, the \(z\) - value (\(z_{\alpha/2}\)) increases. Since \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\) (with \(\sigma\) and \(n\) constant), as \(z_{\alpha/2}\) increases, the margin of error \(E\) increases.

Answer:

(a) Margin of error: \(12.2\), Lower limit: \(125.2\), Upper limit: \(149.6\)
(b) Margin of error: \(14.6\), Lower limit: \(122.8\), Upper limit: \(152.0\)
(c) Margin of error: \(19.1\), Lower limit: \(118.3\), Upper limit: \(156.5\)
(d) As the confidence level increases, the margin of error increases.