You measure 50 textbooks' weights, and find they have a mean weight of 37 ounces. Assume the population standard deviation is 5.2 ounces. Based on this, construct a 90% confidence interval for the true population mean textbook weight. Give your answers as decimals, to two places

90% confidence interval for the true population mean textbook weight is [35.79 ounces , 38.21 ounces].

Step-by-step explanation:

We are given that you measure 50 textbooks' weights, and find they have a mean weight of 37 ounces.

Assume the population standard deviation is 5.2 ounces.

Firstly, the Pivotal quantity for 90% confidence interval for the population mean is given by;

                           P.Q. =    ~ N(0,1)

where, = sample mean weight = 37 ounces

            = population standard deviation = 5.2 ounces

           n = sample of textbooks = 50

            = true population mean textbook weight

Here for constructing 90% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 90% confidence interval for the population mean, is ;

P(-1.645 < N(0,1) < 1.645) = 0.90  {As the critical value of z at 5%

                                            level of significance are -1.645 & 1.645}  

P(-1.645 < < 1.645) = 0.90

P( < < ) = 0.90

P( < < ) = 0.90

90% confidence interval for = [ , ]

                                          = [ , ]

                                          = [35.79 , 38.21]

Therefore, 90% confidence interval for the true population mean textbook weight is [35.79 ounces , 38.21 ounces].

the answer to the expression is 83