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Get Full Access to Statistics For Engineers And Scientists - 4 Edition - Chapter 5.1 - Problem 9e
Get Full Access to Statistics For Engineers And Scientists - 4 Edition - Chapter 5.1 - Problem 9e

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# In a sample of 80 ten-penny nails, the average weight was

ISBN: 9780073401331 38

## Solution for problem 9E Chapter 5.1

Statistics for Engineers and Scientists | 4th Edition

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Problem 9E

Problem 9E

In a sample of 80 ten-penny nails, the average weight was 1.56 g and the standard deviation was 0.1 g.

a. Find a 95% confidence interval for the mean weight of this type of nail.

b. Find a 98% confidence interval for the mean weight of this type of nail.

c. What is the confidence level of the interval (1.54, 1.58)?

d. How many nails must be sampled so that a 95%confidence interval specifies the mean to within ±0.01 d?

e. Approximately how many nails must be sampled so that a 98% confidence interval will specify the mean to within ±0.01 g?

Step-by-Step Solution:
Step 1 of 3

Solution 9E

Step1 of 6:

Let us consider a random variable X it represents the weight of the penny nail. Here random variable X follows normal distribution with mean standard deviation and

n = 80.

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  95% confidence interval for the mean weight of this type of nail.

b). We need to find 98% confidence interval for the mean weight of this type of nail.

c). We need to check what is the confidence level of the interval (1.54, 1.58)?

d). We need to find how many nails must be sampled so that a 95%confidence interval specifies the mean to within ±0.01 d.

e). We need to find how many nails must be sampled so that a 98% confidence interval will specify the mean to within ±0.01 g.

Step2 of 6:

a).

Here we have to find 95%  CI, let us take .

Now,

= 0.025

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 0.9750 value falls, it falls in row 1.9 under column 0.06.

Hence,

95% confidence interval for the mean weight of this type of nail is given by

(1.5380, 1.5819)

Hence,  95% confidence interval for the mean weight of this type of nail is (1.5380, 1.5819).

Step3 of 6:

b).

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

Now,

= 0.01

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 0.9900 value falls, it falls in row 2.3 under column 0.03.

Hence,

98% confidence interval for the mean weight of this type of nail is given by

(1.5341, 1.5858)

Hence,  98% confidence interval for the mean weight of this type of nail is (1.5341, 1.5858).

Step4 of 6:

c).

We have and n = 80.

In a given information we have upper bound 1.58

Now,

1.58 - 1.56 = (0.0111)

0.02 =  (0.0111)

=

= 1.8018

Hence, = 1.8018.

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 1.8 under column 0.00.

= 0.9641

Area to the right of Z is  1 - 0.9641

= 0.0359.

Now,

= 0.0718

Therefore level is (1-) = 1 - 0.0718

= 0.9282.

Hence level is 0.9282.

Step5 of 6:

d).

We know that , s = 0.1 .

= 0.01

= 0.01

= 384.16

385

Hence, n = 385.

Step6 of 6:

e).

We know that , s = 0.1 .

= 0.01

= 0.01

= 542.889

543

Hence, n = 543.

Step 2 of 3

Step 3 of 3

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In a sample of 80 ten-penny nails, the average weight was