In an examination of purchasing patterns of shoppers, a sample of 16 shoppers revealed that they spent, on average, $54 per hour of shopping. Based on previous years, the population standard deviation is thought to be $21 per hour of shopping. Assuming that the amount spent per hour of shopping is normally distributed, find a 90% confidence interval for the mean amount.

So on this case the 90% confidence interval would be given by (45.39;62.61)    

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

represent the sample mean  

population mean (variable of interest)

represent the sample standard deviation

n=16 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

  (1)

Since the Confidence is 0.90 or 90%, the value of and , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that

Now we have everything in order to replace into formula (1):

So on this case the 90% confidence interval would be given by (45.39;62.61)    

irrational, then, just means all the numbers that aren't rational. every integer is a rational number, since each integer n can be written in the form n/1. however, numbers like 1/2, 45454737/2424242, and -3/7 are also rational, since they are fractions whose numerator and denominator are integers.

step-by-step explanation:

33/100 is what i got