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)

b)  

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

Step-by-step 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".  

represent the sample mean  

population mean (variable of interest)  

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:

b) Calculate the confidence interval

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

(1)  

Since the confidence is 0.90 or 90%, the value of and , 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

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

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

e) none of the above because the measure of angle cba is 101°