Respuesta :
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:
- Sample mean of [tex]\overline{x} = 50[/tex].
- Population standard deviation of [tex]\sigma = 2[/tex].
- Sample size of [tex]n = 100[/tex].
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.
A similar problem is given at https://brainly.com/question/25325640
