Respuesta :
Answer:
The 90% confidence interval of the population proportion is (0.6523, 0.7477).
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:
[tex]n = 250, \pi = \frac{X}{n} = \frac{175}{250} = 0.7[/tex]
90% confidence level
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.7 - 1.645\sqrt{\frac{0.7*0.3}{250}} = 0.6523[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.7 + 1.645\sqrt{\frac{0.7*0.3}{250}} = 0.7477[/tex]
The 90% confidence interval of the population proportion is (0.6523, 0.7477).
Tbe confidence interval marked the range of values for which the true population mean is estimated to be given a certain level of confidence. Hence, the confidence interval is (0.6523, 0.7477)
Since the sample size is large enough, we use the the Z distributuon :
Confidence interval is defined thus :
- Mean ± margin of error
Margin of Error :
- [tex] Z = Z* \sqrt{\frac{pq}{n}}[/tex]
- Mean, p = x/n = 175/250 = 0.7
- q = 1 - 0.7 = 0.3
- Zcritical at 90% = Z* = 1.645
Hence,
- Margin of Error = [tex] 1.645 \sqrt{\frac{0.7\times 0.3 }{250}} = 0.0477[/tex]
Lower confidence boundary = 0.7 - 0.0290 = 0.6523
Upper confidence boundary = 0.7 + 0.0290 = 0.7477
Therefore, the confidence interval is (0.6523, 0.7477)
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