Construct a 99% confidence interval for the population mean, mu. Assume the population has a normal distribution. A group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

99% confidence interval for the population mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% confidence interval for the population mean is given by;

         P.Q. = ~

where, = sample mean age of selected students = 22.4 years

             s = sample standard deviation = 3.8 years

             n = sample of students = 19

              = population mean

Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.

So, 99% confidence interval for the population mean, is ;

P(-2.878 < < 2.878) = 0.99  {As the critical value of t at 18 degree of

                                                freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < < 2.878) = 0.99

P( < < ) = 0.99

P( < < ) = 0.99

99% confidence interval for = [ ,  ]

                                                 = [ , ]

                                                 = [19.891 , 24.909]

Therefore, 99% confidence interval for the population mean is [19.891 , 24.909].

answer:      

step-by-step explanation:

b. x + 1

step-by-step explanation:

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