Question

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.

Answer 1

Answer:

$54-1.64\frac{21}{\sqrt{16}}=45.39$    

$54 +1.64\frac{21}{\sqrt{16}}=62.61$    

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".

$\bar X=54$ represent the sample mean  

$\mu$ population mean (variable of interest)

$\sigma=21$ 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:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

Since the Confidence is 0.90 or 90%, the value of $\alpha=1-0.9=0.1$ and $\alpha/2 =0.05$, 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 $z_{\alpha/2}=1.64$

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

$54-1.64\frac{21}{\sqrt{16}}=45.39$    

$54 +1.64\frac{21}{\sqrt{16}}=62.61$    

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