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

Respuesta :

Answer:

a) The point estimate of the population mean is [tex]\hat \mu = \bar X =24.50[/tex]

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".

[tex]\bar X=24.50[/tex] represent the sample mean

[tex]\mu[/tex] 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:

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

The point estimate of the population mean is [tex]\hat \mu = \bar X =24.50[/tex]

3) Part b

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

[tex]df=n-1=20-1=19[/tex]

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], 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 [tex]t_{\alpha/2}=1.73[/tex]

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

[tex]24.5-1.73\frac{3}{\sqrt{20}}=23.339[/tex]    

[tex]24.5+1.73\frac{3}{\sqrt{20}}=25.661[/tex]

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