Question

The research department at the company took a sample of 25 comparable textbooks and collected information on their prices. This information produces a mean price of $145 for this sample. It is known that the standard deviation of the prices of all such textbooks is $35 and the population of such prices is normal. (a) What is the point estimate of the mean price of all such textbooks? (b) Construct a 90% confidence interval for the mean price of all such college textbooks.

Answer 1

a)
`\hat \mu = \bar X = 145`

b)
`145-1.64(35)/(√(25))=133.52`

`145+1.64(35)/(√(25))=156.48`

So on this case the 90% confidence interval would be given by (133.52;156.48)

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=145` represent the sample mean

`\mu` population mean (variable of interest)

`\sigma=35` represent the population standard deviation

n=25 represent the sample size

a) For this case the best point of estimate for the population mean is the sample mean:

`\hat \mu = \bar X = 145`

b) Calculate the confidence interval

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

`\bar X \pm z_(\alpha/2)(\sigma)/(√(n))` (1)

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: "=-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):

`145-1.64(35)/(√(25))=133.52`

`145+1.64(35)/(√(25))=156.48`

So on this case the 90% confidence interval would be given by (133.52;156.48)