Question

College officials want to estimate the percentage of students who carry a gun, knife, or other such weapon. How many randomly selected students must be surveyed in order to be 95% confident that the sample percentage has a margin of error of 1.5 percentage points? (a) Assume that there is no available information that could be used as an estimate of p. Answer: (b) Assume that another study indicated that 6% of college students carry weapons. Answer:

Answer 1

To estimate the percentage of students who carry a weapon with a 95% confidence level and a margin of error of 1.5 percentage points, you can use the formula for sample size calculation. Assuming no available information on the estimated proportion, you would need to survey approximately 1068 students. However, if another study indicated that 6% of college students carry weapons, you would only need to survey approximately 484 students.

Step-by-step explanation:

To determine the sample size needed to estimate the percentage of students who carry a weapon with a 95% confidence level and a margin of error of 1.5 percentage points, we need to use the formula:

n = (Z^2 * p * (1-p)) / (E^2)

Where:

n = sample size

Z = Z-value corresponding to the desired confidence level (95% confidence level corresponds to a Z-value of approximately 1.96)

p = estimated proportion of students carrying a weapon

E = margin of error (1.5 percentage points = 0.015)

(a) If we assume that there is no available information on the estimated proportion of students carrying a weapon, we can use a conservative estimate of p = 0.5 (since this would yield the largest sample size)

n = (1.96^2 * 0.5 * (1-0.5)) / (0.015^2) ≈ 1067.56

So, approximately 1068 students should be surveyed.

(b) If another study indicated that 6% of college students carry weapons, we can use this value as the estimated proportion of students carrying a weapon:

n = (1.96^2 * 0.06 * (1-0.06)) / (0.015^2) ≈ 483.69

So, approximately 484 students should be surveyed.

Answer 2

In this case, assuming a conservative estimate of 0.5 for the proportion of students carrying a weapon, a 95% confidence level, and a margin of error of 1.5 percentage points, the required sample size is 3852 students.

Step-by-step explanation:

To find the sample size required for estimating the percentage of students who carry a weapon, we need to use the formula for sample size calculation:

`n = (Z^2 * p * (1-p)) / E^2`

where:

• n is the required sample size
• Z is the Z-score corresponding to the desired confidence level (in this case, 95% confidence level corresponds to a Z-score of 1.96)
• p is the estimated proportion of students carrying a weapon (in part (a), there is no available estimate of p, so we will use 0.5 as a conservative estimate)
• E is the desired margin of error (in this case, 1.5 percentage points, so we convert it to proportion by dividing it by 100, giving 0.015)

Plugging in the values, we get:

`n = (1.96^2 * 0.5 * (1-0.5)) / (0.015^2)`

n = 3841.78

Since the sample size should be a whole number, we round up to the nearest whole number. Therefore, 3852 randomly selected students must be surveyed in order to be 95% confident that the sample percentage has a margin of error of 1.5 percentage points.