Question

A random sample of 92 sophomores are asked whether they plan to attend Homecoming. Of these, 51 sophomores said they will attend. What is the margin of error at 90% confidence and its interpretation?

0.102; 90% of the time we will capture the true proportion of sophomores who will attend Homecoming within 0.102 of the population proportion

0.085; 90% of the time we will capture the true proportion of sophomores who will attend Homecoming within 0.085 of the population proportion

0.102; we are 90% confident that the true proportion of sophomores who will attend Homecoming is within 0.102 of the sample proportion

0.085; we are 90% confident that the true proportion of sophomores who will attend Homecoming is within 0.085 of the sample proportion

Answer 1

Answer:

D) 0.085; we are 90% confident that the true proportion of sophomores who will attend Homecoming is within 0.085 of the sample proportion.

Step-by-step explanation:

The margin of error for a sample proportion can be calculated using the formula:

$\textsf{Margin of Error} = z \sqrt{\dfrac{\hat{p} (1 - \hat{p})}{n}}$

where:

• z is the z-score associated with the confidence level.
• $\hat{p}$ is the sample proportion.
• n is the sample size.

In this case:

• z = 1.645 (for a 90% confidence level)
• $\hat{p} = \dfrac{51}{92}$
• n = 92

Substitute the values into the formula:

$\textsf{Margin of Error} = 1.645 \times \sqrt{\dfrac{\frac{51}{92} \times \left(1 - \frac{51}{92}\right)}{92}}$

Now, calculate the margin of error:

$\textsf{Margin of Error} = 0.08524348128...$

$\textsf{Margin of Error} = 0.085\; \sf (3\;d.p.)$

So, at 90% confidence, the margin of error is approximately 0.085. The interpretation of this margin of error is that we are 90% confident that the true proportion of sophomores who plan to attend Homecoming is within 0.085 (plus or minus) of the observed proportion.

Answer 2

The correct answer is: 0.085; we are 90% conf.ident that the true proportion of sophomores who will attend Homecom.ing is within 0.085 of the sample proportion.

Let's crunch the numbers:

Margin of er.ror = (Z * √(p * (1 - p) / n))

Given a 90% confidence level, Z is approxima.tely 1.645.

Margin of er.ror = (1.645 * √(0.554 * 0.446 / 92))

Calculating this gives approximately 0.102.

So, the correct interpretation is:

0.102; 90% of the time, we will capture the true proportion of sophomores who will attend Homeco.ming within 0.102 of the population proportion.