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A sample of 87 glass sheets has a mean thickness of 4.20

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

Solution for problem 4SE Chapter 5

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 4SE

A sample of 87 glass sheets has a mean thickness of 4.20 mm with a standard deviation of 0.10 mm.

a. Find a 98% confidence interval for the population mean thickness.

b. What is the level of the confidence interval (4.185, 4.215)?

c. How many glass sheets must be sampled so that a 98% confidence interval will specify the mean to within ±0.015?

Step-by-Step Solution:

Step 1 of 3:

Let n denotes the sample of 87 glass.

Let  denotes the mean thickness is 4.20 mm and

Let s denotes the standard deviation is 0.10 mm.

From the given information we know that n=87, =4.20 mm and s=0.10 mm.

Our goal is:

a). We need to find the 98% confidence interval for the population mean thickness.

b). We need to find the level of the confidence interval (4.185,4.215).

c). We need to find how many glass sheets must be sampled so that a 98% confidence

     interval will specify the mean to within 0.015.

 

a).

Now we need to find the 98% confidence interval for the population mean thickness.

The formula of the confidence interval is

The 98% of confidence interval of  is

                                        

2.33.

Now we have to find the confidence interval.

We know that ,z and n values.

Substitute all the values.

Therefore, the confidence interval is 4.2249 and 4.1751.

Step 2 of 3:

b).

Now we have to find the level of the confidence interval (4.185,4.215).

4.215=

4.215=

.

Then the area of the right angle z is 1-1.40

From the area under the normal curve.

=1-0.9192

=0.0808=.

Then,1-.

=1-2(0.0808)

=1-(0.1616)

=0.8384.

Therefore, the level of the confidence interval is 83.84%.

Step 3 of 3

Chapter 5, Problem 4SE is Solved
Textbook: Statistics for Engineers and Scientists
Edition: 4
Author: William Navidi
ISBN: 9780073401331

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