Question

In an effort to estimate the mean amount spent per customer for dinner at major Atlanta restaurant, data were collected for a sample of 49 customers over a period of three weeks. Assume a population standard deviation of $2.50. If the sample mean is $22.60, compute a 98% confidence interval for population mean?

Answer 1

Answer:

Confidence Interval is (21.76928571,23.43071429)

Step-by-step explanation:

The Formula we are going to use is:

$P=\mu$±$z_{\alpha/2}*\frac{\sigma}{\sqrt{n}}$

Where:

μ is the mean of sample

σ is the standard deviation of population

n is the sample size

z is the distribution

In our case:

μ=$22.60,σ=$2.50

For 98% Confidence level, 1-α=98%

α=2%

α/2=1%=0.01

From Cumulative standardized Normal Distribution table

$z_{\alpha/2}$=2.326

$P=\mu\±z_{\alpha/2}*\frac{\sigma}{\sqrt{n}}\\P=22.60\±2.326*\frac{2.50}{\sqrt{49}}\\P=22.60\±0.8307142857$

P=23.430714      P=21.769286

Confidence Interval is (21.76928571,23.43071429)