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Sagot :
The sample mean weight, from this random sample of 135 pigs, fall within the confidence interval. (Option A: yes)
How to find the confidence interval for large samples (sample size > 30)?
Suppose that we have:
- Sample size n > 30
- Sample mean = [tex]\overline{x}[/tex]
- Sample standard deviation = [tex]s[/tex]
- Population standard deviation = [tex]\sigma[/tex]
- Level of significance = [tex]\alpha[/tex]
Then the confidence interval is obtained as
- Case 1: Population standard deviation is known
[tex]\overline{x} \pm Z_{\alpha /2}\dfrac{\sigma}{\sqrt{n}}[/tex]
- Case 2: Population standard deviation is unknown.
[tex]\overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}[/tex]
For this case, we work with sample standard deviation(you can choose even population standard deviation, it won't matter as both are not given here).
We're provided that:
- Sample size = n = 135
- Confidence interval is of 90%, therefore, Level of significance = [tex]\alpha[/tex] = 100 - 90% = 10% = 10/100 = 0.1 (converted percent to decimal)
- The critical value of Z at 0.1 level of significance is [tex]\pm 1.645[/tex]
- Confidence interval's limits = 75 (lower limit) and 90(upper limit) (in pounds)
- Sample mean = [tex]\overline{x}[/tex]
- Sample standard deviation = [tex]s[/tex]
Since the formula for limits of confidence interval is:
[tex]\overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}} = \overline{x} \pm 1.645\dfrac{s}{\sqrt{135}}[/tex]
That has:
[tex]\text{Lower limit} = 75 = \overline{x} - 1.645\dfrac{s}{\sqrt{135}}\\\\\text{Upper limit} = 90 = \overline{x} +1.645\dfrac{s}{\sqrt{135}}\\\\[/tex]
Thus, we get:
Adding both the equations, we get:
[tex]165 = 2\overline{x}\\\\\overline{x} = 82.5 \: \rm pounds[/tex]
That is between 75 and 90 pounds.
Thus, the sample mean weight, from this random sample of 135 pigs, fall within the confidence interval. (Option A: yes)
Learn more about confidence interval for large samples here:
https://brainly.com/question/13770164
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