Question

fast-food bills for drive-thru customers a random sample of 49 cars in the drive - thru of a popular fast food restaurant revealed an average bill of $19.85 per car. the population standard deviation is $5.83. estimate the mean bill for all cars from the drive - thru with 98% confidence. round intermediate and final answers to two decimal places.
$\square<\mu<\\$\square$

Explanation:

Step1: Find the z - value

For a 98% confidence level, the significance level \(\alpha=1 - 0.98=0.02\). Then \(\alpha/2=0.01\).
Looking up in the standard normal distribution table, \(z_{\alpha/2}=z_{0.01}\approx 2.33\)

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

The formula for the margin of error when the population standard deviation \(\sigma\) is known is \(E = z_{\alpha/2}\times\frac{\sigma}{\sqrt{n}}\)
Given \(n = 49\), \(\sigma=5.83\), \(z_{\alpha/2}=2.33\)
\(E=2.33\times\frac{5.83}{\sqrt{49}}=2.33\times\frac{5.83}{7}\)
\(E = 2.33\times0.8329\approx1.94\)

Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x} + E\)
Given \(\bar{x}=19.85\)
\(19.85- 1.94<\mu<19.85 + 1.94\)

Answer:

\(17.91<\mu<21.79\)