A random sample of 8 fields of barley has a mean yield of 49.6 bushels per acre and standard deviation of 2.33 bushels per acre. Determine the 99% confidence interval for the true mean yield. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

99% confidence interval for the true mean yield is [46.718 , 52.482].

Step-by-step explanation:

We are given that a random sample of 8 fields of barley has a mean yield of 49.6 bushels per acre and standard deviation of 2.33 bushels per acre.

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

         P.Q. = ~

where, = sample mean yield = 49.6 bushels per acre

             s = sample standard deviation = 2.33 bushels per acre

             n = sample of fields of barley = 8

              = true 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 true mean, is ;

P(-3.499 < < 3.499) = 0.99  {As the critical value of t at 7 degree of

                                                freedom are -3.499 & 3.499 with P = 0.5%}

P(-3.499 < < 3.499) = 0.99

P( < < ) = 0.99

P( < < ) = 0.99

99% confidence interval for = [ ,  ]

                                                 = [ , ]

                                                 = [46.718 , 52.482]

Therefore, 99% confidence interval for the true mean yield is  [46.718 , 52.482].

17   x 256=4352

sorry if this is wrong, i'm new.

step-by-step explanation: