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While doing an article on the high cost of college education, a reporter took a random sample of the cost of new textbooks for a semester. Her sample consisted of 20 students with a mean of $212.45 and a standard deviation of $38.92. A. Find the 99% confidence interval estimate of the true mean textbook cost for the semester based on her sample. Assume the cost of new textbooks is normally distributed.B. Based on part a, is an average cost of $200 for a semester a likely value for the population mean? Why or why not?

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Answer:

The answer is below

Step-by-step explanation:

Given that:

mean (μ) = $212.45, standard deviation (σ) = $38.92 and sample size (n) = 20 students.

a) Confidence (C) = 99% = 0.99

α = 1 - C = 1 - 0.99 = 0.01

α/2 = 0.01/2 = 0.005

The z score of α/2 is the same as the z score of 0.495 (0.5 - 0.005) which is equal to 2.576. This means that Z(α/2) = 2.576

The margin of error (E) is given as:

[tex]E=Z_{\frac{\alpha}{2} }*\frac{\sigma}{\sqrt{n} } \\\\Substituting:\\\\E=2.576*\frac{38.92}{\sqrt{20} } \\\\E=8.7[/tex]

The confidence interval = (μ ± E) = (212.45 ± 8.7) = (203.75, 221.15)

Hence the 99% confidence interval is between 203.75 and 221.15.

b) The population mean is the same as the sample mean. Hence the population mean = $212.45