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Sagot :
Using the z-distribution, it is found that the 95% confidence interval for the true proportion of adults who said that they will travel more this year is (0.347, 0.433).
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]\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 z-score that has a p-value of [tex]\frac{1+\alpha}{2}[/tex].
195 out of 500 people said they would take more vacations this year than last year, hence:
[tex]n = 500, \pi = \frac{195}{500} = 0.39[/tex]
95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{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}} = 0.39 - 1.96\sqrt{\frac{0.39(0.61)}{500}} = 0.347[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.39 + 1.96\sqrt{\frac{0.39(0.61)}{500}} = 0.433[/tex]
The 95% confidence interval for the true proportion of adults who said that they will travel more this year is (0.347, 0.433).
A similar problem is given at https://brainly.com/question/15850972
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