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

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)

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)(s)/(√(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(3)/(√(20))=23.339`

`24.5+1.73(3)/(√(20))=25.661`

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