Question

The owner of the Peach Pit restaurant examined 61 randomly selected sample receipts for parties of four and writes down how much they spent. The sample average equals to $39.20 with a sample standard deviation of $18.35. Construct a 95% confidence interval for the mean of the amount spent by all parties of four at Peach Pit.Also find the margin of error. Round your answers to 2 decimal places.

Answer 1

The formula for confidence interval of population mean is

where

= sample mean

s = sample standard deviation

n = sample size

= critical z-value

The "Z" values for different Confidence Interval are listed below:

Confidence Interval	Z
80%	1.282
85%	1.440
90%	1.645
95%	1.960
99%	2.576
99.5%	2.807
99.9%	3.291

Given

n = 61

From the table, the Z-value for 95% confidence level is 1.96

s = 18.35

Now

The 95% confidence interval for population mean is

  

( 34.60 , 43.80 )

Therefore,

the 95% confidence interval for the mean of the amount spent by all parties of four at Peach Pit is ( 34.60 , 43.80 ).

Margin of error = critical value * standard error

  

  

Therefore, margin of error  .