The Highway Safety Department wants to study the driving habits of individuals. A sample of 37 cars traveling on a particular stretch of highway revealed an average speed of 70.7 miles per hour with a standard deviation of 6.3 miles per hour. Round to 4 decimal places. 1.Calculate a 90% confidence interval for the true mean speed of all cars on this particular stretch of highway

90% confidence interval for the true mean speed of all cars on this particular stretch of highway is [68.9517 miles per hour , 72.4483 miles per hour].

Step-by-step explanation:

We are given that a sample of 37 cars traveling on a particular stretch of highway revealed an average speed of 70.7 miles per hour with a standard deviation of 6.3 miles per hour.

Firstly, the pivotal quantity for 90% confidence interval for the true mean is given by;

                            P.Q. =  ~

where, = sample average speed of cars = 70.7 miles per hour

             s = sample standard deviation = 6.3 miles per hour

             n = sample of cars = 37

              = true mean speed

Here for constructing 90% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.

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

P(-1.688 < < 1.688) = 0.90  {As the critical value of t at 36 degree of

                                 freedom are -1.688 & 1.688 with P = 5%}  

P(-1.688 < < 1.688) = 0.90

P( < < ) = 0.90

P( < < ) = 0.90

90% confidence interval for = [ , ]

                    = [ , ]

                    = [68.9517 miles per hour , 72.4483 miles per hour]

Therefore, 90% confidence interval for the true mean speed of all cars on this particular stretch of highway is [68.9517 miles per hour , 72.4483 miles per hour].

The interpretation of the above interval is that we are 90% confident that the true mean speed of all cars will lie between 68.9517 miles per hour and 72.4483 miles per hour.

(c, 0) i might be wrong so don't put full faith on my answer