In a sample of 100 steel wires the average breaking strength is 50 kN, with a standard deviation of 2 kN. a. Find a 95% confidence interval for the mean breaking strength of this type of wire. b. Find a 99% confidence interval for the mean breaking strength of this type of wire. c. An engineer claims that the mean breaking strength is between 49.7 kN and 50.3 kN. With what level of confidence can this statement be made? d. How many wires must be sampled so that a 95% confidence interval specifies the mean breaking strength to within ±0.3 kN? e. How many wires must be sampled so that a 99% confidence interval specifies the mean breaking strength to within ±0.3 kN?

Answer:

.

Step-by-step explanation:

1.

Alpha = 0.05. alpha/2 = 0.025

Under the z table

Z 0.025 = 1.96

50+-1.96(2/√100)

= 50+-0.39

50+0.39 = 50.39

50-0.39 = 49.61

2.

At 99%

Alpha = 0.01

Alpha/2 = 0.005

In z table

= 2.58

50+-(2.58)(2√100)

= 50 +- 0.52

= 50+0.52 = 50.52

50 - 52 = 49.48

Please check attachment for answers 3 to 5. The test editor did not work properly. It kept flagging some of my typed content.

Using the z-distribution, it is found that:

a) The 95% confidence interval for the mean breaking strength of this type of wire, in kN, is (49.61, 50.39).

b) The 99% confidence interval for the mean breaking strength of this type of wire, in kN, is (49.49, 50.52).

c) The confidence level is of 86.64%.

d) 171 wires must be sampled.

e) 295 wires must be sampled.

We are given the standard deviation for the population, which is why the z-distribution is used to solve this question.

The information given is:

The confidence interval is:

[tex]\overline{x} \pm z\frac{\sigma}{\sqrt{n}}[/tex]

The margin of error is of:

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

Item a:

The first step is finding the critical value, which is z with a p-value of [tex]\frac{1 + \alpha}{2}[/tex], in which [tex]\alpha[/tex] is the confidence level.

In this problem, [tex]\alpha = 0.95[/tex], thus, z with a p-value of [tex]\frac{1 + 0.95}{2} = 0.975[/tex], which means that it is z = 1.96.

Hence, the interval is:

[tex]\overline{x} - z\frac{\sigma}{\sqrt{n}} = 50 - 1.96\frac{2}{\sqrt{100}} = 49.61[/tex]

[tex]\overline{x} + z\frac{\sigma}{\sqrt{n}} = 50 + 1.96\frac{2}{\sqrt{100}} = 50.39[/tex]

The 95% confidence interval for the mean breaking strength of this type of wire, in kN, is (49.61, 50.39).

Item b:

[tex]\alpha = 0.99[/tex], thus, z with a p-value of [tex]\frac{1 + 0.99}{2} = 0.995[/tex], which means that it is z = 2.575.

[tex]\overline{x} - z\frac{\sigma}{\sqrt{n}} = 50 - 2.575\frac{2}{\sqrt{100}} = 49.49[/tex]

[tex]\overline{x} + z\frac{\sigma}{\sqrt{n}} = 50 + 2.575\frac{2}{\sqrt{100}} = 50.52[/tex]

The 99% confidence interval for the mean breaking strength of this type of wire, in kN, is (49.49, 50.52).

Item c:

The margin of error is:

[tex]M = \frac{50.3 - 49.7}{2} = 0.3[/tex]

Hence:

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

[tex]0.3 = z\frac{2}{\sqrt{100}}[/tex]

[tex]0.2z = 0.3[/tex]

[tex]z = 1.5[/tex]

z = 1.5 has a p-value of 0.9332, hence:

[tex]\frac{1 + \alpha}{2} = 0.9332[/tex]

[tex]1 + \alpha = 2(0.9332)[/tex]

[tex]\alpha = 0.8664[/tex]

The confidence level is of 86.64%.

Item d:

This is n for which M = 0.3, with z = 1.96, hence:

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

[tex]0.3 = 1.96\frac{2}{\sqrt{n}}[/tex]

[tex]0.3\sqrt{n} = 1.96(2)[/tex]

[tex]\sqrt{n} = \frac{1.96(2)}{0.3}[/tex]

[tex](\sqrt{n})^2 = \left(\frac{1.96(2)}{0.3}\right)^2[/tex]

[tex]n = 170.7[/tex]

Rounding up, 171 wires must be sampled.

Item e:

Now z = 2.575, hence:

[tex](\sqrt{n})^2 = \left(\frac{2.575(2)}{0.3}\right)^2[/tex]

[tex]n = 294.7[/tex]

Rounding up, 295 wires must be sampled.

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