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length of growing seasons the growing seasons for a random sample of 44…

Question

length of growing seasons the growing seasons for a random sample of 44 u.s. cities were recorded, yielding a sample mean of 191.8 days and the population standard deviation of 55.8 days. estimate the true population mean of the growing season with 95% confidence. round your answers to at least one decimal place.

$\square < \mu < \square$

Explanation:

Step1: Find the z - value

For a 95% confidence interval, the significance level \(\alpha=1 - 0.95=0.05\). Then \(\alpha/2=0.025\). Looking up in the standard normal distribution table, \(z_{\alpha/2}=z_{0.025} = 1.96\).

Step2: Calculate the margin of error \(E\)

The formula for the margin of error when the population standard deviation \(\sigma\) is known is \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\).
Given \(n = 44\), \(\sigma=55.8\), \(z_{\alpha/2}=1.96\).

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Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu <\bar{x}+E\), where \(\bar{x} = 191.8\).

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Answer:

\(175.3<\mu<208.3\)