Question

A random sample of 20 shoppers was taken. Sample data showed that they spend an average of $24.50 per visit at the Sunday Morning Bookstore. The standard deviation of the sample is $3.00. a) Find a point estimate of the population mean. b) Find the 90% confidence interval of the true mean

Answer 1

Answer:

a) The point estimate of the population mean is $\hat \mu = \bar X =24.50$

b) The 90% confidence interval would be given by (23.339;25.661)    

Step-by-step explanation:

1) 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".

$\bar X=24.50$ represent the sample mean

$\mu$ population mean (variable of interest)

s=3.00 represent the sample standard deviation

n=20 represent the sample size  

2) Part a

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

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

The point estimate of the population mean is $\hat \mu = \bar X =24.50$

3) Part b

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=20-1=19$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ 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: "=-T.INV(0.05,19)".And we see that $t_{\alpha/2}=1.73$

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

$24.5-1.73\frac{3}{\sqrt{20}}=23.339$    

$24.5+1.73\frac{3}{\sqrt{20}}=25.661$

So on this case the 90% confidence interval would be given by (23.339;25.661)