A marketing research company desires to know the mean consumption of meat per week among people over age 30. A sample of 2092 people over age 30 was drawn and the mean meat consumption was 3 pounds. Assume that the population standard deviation is known to be 1.4 pounds. Construct the 95% confidence interval for the mean consumption of meat among people over age 30. Round your answers to one decimal place.

95% confidence interval for the mean consumption of meat among people over age 30 is [2.9 pounds , 3.1 pounds].

Step-by-step explanation:

We are given that a sample of 2092 people over age 30 was drawn and the mean meat consumption was 3 pounds. Assume that the population standard deviation is known to be 1.4 pounds.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

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

where, = sample mean meat consumption = 3 pounds

             s = population standard deviation = 1.4 pounds

            n = sample of people = 2092

             = population mean consumption of meat

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean, is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                  of significance are -1.96 & 1.96}  

P(-1.96 < < 1.96) = 0.95

P( < < ) = 0.95

P( < < ) = 0.95

95% confidence interval for = [ , ]

                 = [ , ]

                 = [2.9 pounds , 3.1 pounds]

Therefore, 95% confidence interval for the mean consumption of meat among people over age 30 is [2.9 pounds , 3.1 pounds].

answer: i think one of the answer is $150 and $300?

step-by-step explanation:

$6.42

step-by-step explanation: