A random sample of 49 text books purchased at a local bookstore showed an average price of $122 with a population standard deviation of $15. Let u (new) be the true mean cost of a text book sold by this store. Construct a confidence interval with a 90% degree of confidence. Clearly label the following: a. Point estimate b. Critical value, c. Margin of error d. Confidence interval e. Interpretation (confidence statement).

Answer:point estimate is $122critical value for the 90% confidence level (1.645)margin of error is $3.52590% confidence interval is $122±3.525there is 90% probability that true population average price of text books is in the range $122±$3.525Step-by-step explanation:Confidence Interval can be calculated using P±ME where P is the point estimate for the mean cost of a text book ( $122 )ME is the margin of error from the mean And margin of error (ME) can be calculated using the formulaME=[tex]\frac{z*s}{\sqrt{N} }[/tex] where z is the critical value for the 90% confidence level (1.645)s is the population standard deviation ($15) N is the sample size (49)Margin of error, ME=[tex]\frac{1.645*15}{\sqrt{49} }[/tex] = 3.525Then 90% confidence interval is $122±3.525To interpret this, there is 90% probability that true population average price of text books is in the range $122±$3.525