I have no idea how to solve this.

In a sample of 16 students, 15 were right-handed. Can we construct a 95% confidence interval for the proportion of all students who are right-handed?

This is a yes/no question.

Since 15/16 is 0.9375, you cannot make a 95% confidence interval according to WebAssign.

I wish I knew why too.

To construct a confidence interval for a proportion, we can use the formula:

Confidence Interval = Sample Proportion ± Margin of Error

First, we need to find the sample proportion. In this case, out of 16 students, 15 were right-handed. Therefore, the sample proportion of right-handed students is 15/16.

Next, we need to calculate the margin of error. The margin of error depends on the desired confidence level, which is given as 95%. For a 95% confidence level, we can use the standard z-value of 1.96, which corresponds to a two-tailed test.

The formula for the margin of error is:

Margin of Error = z * (sqrt(p * (1 - p) / n))

where:
- z is the z-value corresponding to the desired confidence level (1.96 for a 95% confidence level)
- p is the sample proportion of right-handed students
- n is the sample size

Plugging in the values, we get:

Margin of Error = 1.96 * (sqrt((15/16) * (1 - 15/16) / 16))

Finally, we can calculate the confidence interval by subtracting and adding the margin of error from the sample proportion:

Confidence Interval = (15/16) ± Margin of Error

This will give you the lower and upper limits of the confidence interval for the proportion of all students who are right-handed.