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Refer to Exercise 6.a. Find a 98% lower confidence bound

Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi ISBN: 9780073401331 38

Solution for problem 13E Chapter 5.1

Statistics for Engineers and Scientists | 4th Edition

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Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi

Statistics for Engineers and Scientists | 4th Edition

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Problem 13E

Problem 13E

Refer to Exercise 6.

a. Find a 98% lower confidence bound for the mean time.

b. Someone says that the mean time is greater than 134.3 minutes. With what level of confidence, can this statement be made?

Step-by-Step Solution:
Step 1 of 3

Solution 13E

Step1 of 3:

Let us consider a random variable X it represents the surgery time. Here random variable X follows normal distribution with mean standard deviation and n = 123.

That is

                       

The probability density function of normal distribution is given by

, .

Where,

x = random variable

= mean of X

= variance of X

= standard deviation os X

= mathematical constant and its value is 3.14

Here our goal is:

a). We need to find a 98% lower confidence bound for the mean time.

b). We need to find level of confidence, can this statement be made, by taking the mean time is greater than 134.3 minutes.


Step2 of 3:

a).

Here we have to find 98%  CI, let us take .

Now,

 

   = 0.02

Z-scores(are to the right) is given by

                                                                       

is obtained from standard normal table(area under normal curve). In standard normal table we have to see where 0.98 value falls, it falls in row 2.0 under column 0.05.

Hence,

Now 98% lower confidence bound for the mean time is given by

       

     

         

132.7227

Hence, 98% lower confidence bound for the mean time is 132.7227.


Step3 of 3:

b).

We have and n = 123.

In a given information we have lower bound 134.3

Now,

               

                         

     134.3 - 136.9 = (2.0377)

                 -2.6 =  (2.0377)

                                                       =

         = -1.28

Hence, = -1.28.

Z-scores is given by , and this value is obtained from standard normal table(area under normal curve). In standard normal table we have to see in row -1.2 under column 0.08.

 = 0.1003

Area to the right of Z is  1 - 0.1003

                                     = 0.8997.

Hence level is 0.8997.


Step 2 of 3

Chapter 5.1, Problem 13E is Solved
Step 3 of 3

Textbook: Statistics for Engineers and Scientists
Edition: 4
Author: William Navidi
ISBN: 9780073401331

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Refer to Exercise 6.a. Find a 98% lower confidence bound