Respuesta :
Answer:
a) 0.119 (11.9%)
b) 0.261 (26.1%)
c) Yes because 0.2 is inside the interval.
d) n = 2,554
Step-by-step explanation:
We have to calculate a 99% confidence interval for the proportion.
The sample proportion is p=0.19.
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The standard error of the proportion is:
[tex]\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.19*0.81}{200}}\\\\\\ \sigma_p=\sqrt{0.00077}=0.03[/tex]
The critical z-value for a 99% confidence interval is z=2.576.
The margin of error (MOE) can be calculated as:
[tex]MOE=z\cdot \sigma_p=2.576 \cdot 0.03=0.071[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=p-z \cdot \sigma_p = 0.19-0.071=0.119\\\\UL=p+z \cdot \sigma_p = 0.19+0.071=0.261[/tex]
The 99% confidence interval for the population proportion is (0.119, 0.261).
c) The value 0.2 is included in the interval. This means that is one of the values that, with 99% confidence, the true proportion can take.
d) We have to calculate the minimum sample size n needed to have a margin of error below 0.02.
To do that, we use the margin of error formula in function of n:
[tex]MOE=z\cdot \sqrt{\dfrac{p(1-p)}{n}}\\\\n=\left(\dfrac{z}{MOE}\right)^2\cdot p(1-p)\\\\\\n=\left(\dfrac{2.576}{0.02}\right)^2\cdot 0.19(1-0.19)\\\\\\n=(128.8)^2\cdot0.19\cdot 0.81\\\\\\n=16,589.44\cdot 0.1539\\\\\\n=2,553.115\approx2,554[/tex]
The minimum sample size to have this margin of error is n = 2,554.
