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Sagot :
Answer:
The 95% confidence interval for the proportion of cans in the shipment that meet the specification is (0.741, 0.917).
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].
For this problem, we have that:
70 cans, 58 meet the specification for puncture resistance. This means that [tex]n = 50, \pi = \frac{58}{70} = 0.829[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.829 - 1.96\sqrt{\frac{0.829*0.171}{70}} = 0.741[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.829 + 1.96\sqrt{\frac{0.829*0.171}{70}} = 0.917[/tex]
The 95% confidence interval for the proportion of cans in the shipment that meet the specification is (0.741, 0.917).
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