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?

Confidence Interval is (21.76928571,23.43071429)

Step-by-step explanation:

The Formula we are going to use is:

±

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

=2.326

P=23.430714      P=21.769286

Confidence Interval is (21.76928571,23.43071429)

answer: 1,380

step-by-step explanation:

1: you turn %15 into a decimal by moving the decimal point two places to the left.

2: and then you multiply .15 by 1,200 , which will get you 180.00

3: from there you will add 180 to 1,200

4: which will get you 1,380

hope it *im kind of slow so idk if my answer is right*

50x5=$250

step-by-step explanation: