Respuesta :
The sample size of the confidence interval test of a proportion is considered appropriately large if np and n(1 - p) > 5.
Given that Arthur is conducting a study on the preferred study options of students from East County College and that 25% of those surveyed preferred studying abroad. Thus, p = 25% = 0.25 and 1 - 0.25 = 0.75
np = 32(0.25) = 8
n(1 - p) = 32(0.75) = 24
Thus, the sample size is appropriately large.
The margin of error is given by:
[tex]\pm z_{ \alpha /2} \sqrt{ \frac{p(1-p)}{n} } [/tex]
The [tex]z_{ \alpha /2}[/tex] for 95% confidence interval is 1.96
Thus, the margin of error is given by:
[tex]\pm1.96 \sqrt{ \frac{0.25(1-0.25)}{32} } =\pm1.96 \sqrt{\frac{0.25(0.75)}{32}} \\ \\ =\pm1.96 \sqrt{ \frac{0.1875}{32} } =\pm1.96 \sqrt{0.005859375} \\ \\ =\pm1.96(0.0765)=\pm0.15[/tex]
Therefore, the statement that holds true is "as the sample size is appropriately large, the margin of error is ±0.15".
Given that Arthur is conducting a study on the preferred study options of students from East County College and that 25% of those surveyed preferred studying abroad. Thus, p = 25% = 0.25 and 1 - 0.25 = 0.75
np = 32(0.25) = 8
n(1 - p) = 32(0.75) = 24
Thus, the sample size is appropriately large.
The margin of error is given by:
[tex]\pm z_{ \alpha /2} \sqrt{ \frac{p(1-p)}{n} } [/tex]
The [tex]z_{ \alpha /2}[/tex] for 95% confidence interval is 1.96
Thus, the margin of error is given by:
[tex]\pm1.96 \sqrt{ \frac{0.25(1-0.25)}{32} } =\pm1.96 \sqrt{\frac{0.25(0.75)}{32}} \\ \\ =\pm1.96 \sqrt{ \frac{0.1875}{32} } =\pm1.96 \sqrt{0.005859375} \\ \\ =\pm1.96(0.0765)=\pm0.15[/tex]
Therefore, the statement that holds true is "as the sample size is appropriately large, the margin of error is ±0.15".
Answer:
as the sample size is approximately large, the margin of error is 0.15