Question

length of growing seasons the growing seasons for a random sample of 40 u.s. cities were recorded, yielding a sample mean of 192.5 days and the population standard deviation of 53.5 days. estimate the true population mean of the growing season with 99% confidence. round your answers to at least one decimal place.

Explanation:

Step1: Find the critical value \( z_{\alpha/2} \)

For a \( 99\% \) confidence interval, \( \alpha=1 - 0.99 = 0.01 \), so \( \alpha/2=0.005 \).
From the standard normal distribution table, \( z_{0.005}=2.576 \).

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 \( n = 40 \), \( \sigma=53.5 \), and \( z_{\alpha/2}=2.576 \).
\( E=2.576\times\frac{53.5}{\sqrt{40}} \)
First, calculate \( \sqrt{40}\approx6.3246 \)
\( E = 2.576\times\frac{53.5}{6.3246}\approx2.576\times8.46\approx21.8 \)

Step3: Calculate the confidence interval

The confidence interval for the population mean \( \mu \) is \( \bar{x}-E<\mu<\bar{x} + E \)
Given \( \bar{x}=192.5 \)
\( 192.5-21.8<\mu<192.5 + 21.8 \)

Answer:

\( 170.7<\mu<214.3 \)