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Sagot :
Answer:
The 98% confidence interval for the population proportion of disks which are defective is (0.082, 0.118).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
Suppose a sample of 1536 floppy disks is drawn. Of these disks, 1383 were not defective.
1536 - 1383 = 153
This means that [tex]n = 1536, \pi = \frac{153}{1536} = 0.1[/tex]
98% confidence level
So [tex]\alpha = 0.02[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.02}{2} = 0.99[/tex], so [tex]Z = 2.327[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1 - 2.327\sqrt{\frac{0.1*0.9}{1536}} = 0.082[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1 + 2.327\sqrt{\frac{0.1*0.9}{1536}} = 0.118[/tex]
The 98% confidence interval for the population proportion of disks which are defective is (0.082, 0.118).
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