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sergey [27]
3 years ago
8

The Highway Safety Department wants to study the driving habits of individuals. A sample of 37 cars traveling on a particular st

retch 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
Mathematics
1 answer:
kodGreya [7K]3 years ago
8 0

Answer:

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. = \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

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

             s = sample standard deviation = 6.3 miles per hour

             n = sample of cars = 37

             \mu = true mean speed

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

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

P(-1.688 < t_3_6 < 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 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.688) = 0.90

P( -1.688 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 1.688 \times {\frac{s}{\sqrt{n} } } ) = 0.90

P( \bar X-1.688 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+1.688 \times {\frac{s}{\sqrt{n} } } ) = 0.90

<em><u>90% confidence interval for</u></em> \mu = [ \bar X-1.688 \times {\frac{s}{\sqrt{n} } } , \bar X+1.688 \times {\frac{s}{\sqrt{n} } } ]

                    = [ 70.7-1.688 \times {\frac{6.3}{\sqrt{37} } } , 70.7+1.688 \times {\frac{6.3}{\sqrt{37} } } ]

                    = [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].

<em>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.</em>

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