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.

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=145[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

[tex]\sigma=35[/tex] 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:

[tex]\hat \mu = \bar X = 145[/tex]

b) Calculate the confidence interval

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

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

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: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]

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

[tex]145-1.64\frac{35}{\sqrt{25}}=133.52[/tex]

[tex]145+1.64\frac{35}{\sqrt{25}}=156.48[/tex]

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