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Sagot :
Answer:
The sample proportion is 0.88.
The margin of error is of 0.04.
The interval likely to contain the true population proportion is (0.84,0.92).
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 z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
Margin of error:
The margin of error is of:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
Sample of 230 teenagers, 202 would like to have smaller class sizes at their school.
This means that [tex]n = 230[/tex], and that the sample proportion is:
[tex]\pi = \frac{202}{230} = 0.88[/tex]
The sample proportion is 0.88.
Standard 95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
Margin of error:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]M = 1.96\sqrt{\frac{0.88*0.12}{230}}[/tex]
[tex]M = 0.04[/tex]
The margin of error is of 0.04.
Confidence interval:
Sample proportion plus/minus margin of error. So
0.88 - 0.04 = 0.84
0.88 + 0.04 = 0.92
The interval likely to contain the true population proportion is (0.84,0.92).
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