Respuesta :
Answer:
0.12 ± 1.96 * √(0.12(0.88) / 100)
Step-by-step explanation:
Confidence interval :
Phat ± Zcritical * √(phat(1 -phat) / n)
Phat = 12/100 = 0.12
1 - phat = 0.88
Zcritical at 95% = 1.96
Hence, we have :
0.12 ± 1.96 * √(0.12(0.88) / 100)
0.12 ± 1.96 * 0.0324961
0.12 ± 0.0636924
Lower boundary = (0.12 - 0.0636924) = 0.0563
Upper boundary = 0.12 + 0.0636924 = 0.1837
The calculations for 95% confidence interval for this case is given as: [tex]CI = 0.12 \pm 1.96 \times \sqrt{\dfrac{0.12(1 - 0.12)}{100}}[/tex]
How to find the confidence interval for proportion with large sample ?
For large enough sample( size > 30), let the population proportion of a quantity be denoted by random variable [tex]p[/tex]
Then, we get:
[tex]p \sim N(\hat{p}, \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}})[/tex]
where
[tex]\hat{p}[/tex] = estimated(mean value) proportion of that quantity
and n = size of sample drawn.
It is visible that as we increase the value of n, the standard deviation decreases, therefore, forcing the values of population proportion to be closer to the estimated proportion.
Margin of error is the distance between the mean and one of the end point of the confidence interval(assuming its equal on both the sides of the mean). The margin of error with level of significance [tex]\alpha[/tex] is calculated as:
[tex]MOE = Z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
where [tex]Z_{\alpha/2}[/tex] is the critical value of the test statistic for level of significance [tex]\alpha[/tex]
The confidence interval is then calculated as:
[tex]CI = \hat{p} \pm MOE[/tex]
For the considered case,
the confidence interval is to be found for 95% confidence, therefore,
level of significance = 100% - 95% = 5% = 0.05
- At this level of significance, the critical value of test statistic Z is 1.96
- The sample size given is n = 100
The estimated proportion of defective chips estimated from sample is:
- [tex]\hat{p}[/tex] = Fraction of defective chips to total chips = 12/100 = 0.12
Thus, we get the confidence interval for this case as:
[tex]CI = \hat{p} \pm Z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\CI = 0.12 \pm 1.96 \times \sqrt{\dfrac{0.12(1 - 0.12)}{100}}[/tex]
Thus, the calculations for 95% confidence interval for this case is given as: [tex]CI = 0.12 \pm 1.96 \times \sqrt{\dfrac{0.12(1 - 0.12)}{100}}[/tex]
Learn more about confidence interval for sample proportion here:
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