Respuesta :
Answer:
The 95% confidence interval is [tex] 10.91 < \mu <11.28 [/tex]
The 99% confidence interval is [tex] 10.85 < \mu <11.33 [/tex]
Step-by-step explanation:
From the question we are told that
The sample mean is [tex]\= x = 11.09 \ tons[/tex]
The standard deviation is [tex]\sigma = 0.73 \ tons[/tex]
The sample size is n =60
From the question we are told the confidence level is 95% , hence the level of significance is
[tex]\alpha = (100 - 95 ) \%[/tex]
=> [tex]\alpha = 0.05[/tex]
Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }[/tex]
=> [tex]E = 1.96 * \frac{ 0.73 }{\sqrt{n60 }[/tex]
=> [tex]E = 0.1847 [/tex]
Generally 95% confidence interval is mathematically represented as
[tex]\= x -E < \mu < \=x +E[/tex]
=> [tex] 11.09 - 0.1847 < \mu <11.09 + 0.1847 [/tex]
=> [tex] 10.91 < \mu <11.28 [/tex]
From the question we are told the confidence level is 99% , hence the level of significance is
[tex]\alpha = (100 - 99 ) \%[/tex]
=> [tex]\alpha = 0.01/tex]
Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 2.58 [/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }[/tex]
=> [tex]E = 2.58 * \frac{ 0.73 }{\sqrt{n60 }[/tex]
=> [tex]E = 0.2431 [/tex]
Generally 95% confidence interval is mathematically represented as
[tex]\= x -E < \mu < \=x +E[/tex]
=> [tex] 11.09 - 0.2431< \mu <11.09 + 0.2431 [/tex]
=> [tex] 10.85 < \mu <11.33 [/tex]
The confidence limits for the mean of the maximum loads of all cables produced by the company are: for 95%, it is [10.935,11.245]. For 99%, it is [10.87 ,11.309].
What is the margin of error for large samples?
Suppose that we have:
- Sample size n > 30
- Sample standard deviation = s
- Population standard deviation = [tex]\sigma[/tex]
- Level of significance = [tex]\alpha[/tex]
Then the margin of error(MOE) is obtained as
- Case 1: Population standard deviation is known
Margin of Error = [tex]MOE = Z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
- Case 2: Population standard deviation is unknown
[tex]MOE = Z_{\alpha/2}\dfrac{s}{\sqrt{n}}[/tex]
where [tex]Z_{\alpha/2}[/tex] is critical value of the test statistic at level of significance
The confidence interval for specific level of significance [tex]\alpha[/tex] is:
[tex][\overline{x} - |MOE|, \overline{x} + |MOE|][/tex]
where [tex]\overline{x}[/tex] is the sample mean.
For this case, we are given that:
- Sample size = n = 60
- Sample mean = [tex]\overline{x}[/tex] = 11.09 tons
- Sample standard deviation = s = 0.73 tons
Calculating the confidence limits of given confidence level:
- Case 1: Confidence interval of 95%
The level of significance = [tex]\alpha = 100 - 95\% = 5\% = 0.05[/tex]
The critical value of Z at this level of significance is [tex]Z_{\alpha/2} = Z_{0.05/2} = \pm 1.645[/tex]
Thus, the confidence interval is:
[tex][\overline{x} - |MOE|, \overline{x} + |MOE|] = [11.09 - 1.645 \times \dfrac{0.73}{\sqrt{60}}, 11.09 + 1.645 \times \dfrac{0.73}{\sqrt{60}}]\\\\\approx [11.09 - 0.155, 11.09 + 0.155] = [10.935,11.245][/tex]
Thus, the confidence interval at 95% is [10.935,11.245]
- Case 2: Confidence interval of 99%
The level of significance = [tex]\alpha = 100 - 99\% = 1\% = 0.01[/tex]
The critical value of Z at this level of significance is [tex]Z_{\alpha/2} = Z_{0.05/2} \approx \pm 2.33[/tex]
Thus, the confidence interval is:
[tex][\overline{x} - |MOE|, \overline{x} + |MOE|] = [11.09 - 2.33 \times \dfrac{0.73}{\sqrt{60}}, 11.09 + 2.33 \times \dfrac{0.73}{\sqrt{60}}]\\\\\approx [11.09 - 0.1847, 11.09 + 0.1847] = [10.87 ,11.31][/tex]
Thus, the confidence interval at 99% is [10.87 ,11.309]
Thus, the confidence limits for the mean of the maximum loads of all cables produced by the company are: for 95%, it is [10.935,11.245]. For 99%, it is [10.87 ,11.309].
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