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Sagot :
Answer:
a) 0.6032
b)
Lower limit: 0.48
Upper limit: 0.72
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].
Question a:
In a random sample of 63 professional actors, it was found that 38 were extroverts.
We use this to find the sample proportion, which is the point estimate for p. So
[tex]\pi = \frac{38}{63} = 0.6032[/tex]
Question b:
Sample of 63 means that [tex]n = 63[/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}} = 0.6032 - 1.96\sqrt{\frac{0.6032*0.3968}{63}} = 0.4824[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6032 + 1.96\sqrt{\frac{0.6032*0.3968}{63}} = 0.724[/tex]
Rounding to two decimal places:
Lower limit: 0.48
Upper limit: 0.72
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