Question

The Community College Survey of Student Engagement reports that 46% of the students surveyed rarely or never use peer or other tutoring resources. Suppose that in reality 40% of community college students never use tutoring services available at their college. In a simulation, we select random samples from a population in which 40% do not use tutoring. For each sample, we calculate the proportion who do not use tutoring. If we randomly sample 100 students from this population, the standard error is approximately 5%. Would it be unusual to see 46% who do not use tutoring in a random sample of 100 students?a) Yes, this would be unusual because 46% is more than one standard error from the mean. It is very rare for a sample to be more than one standard error from mean. b) Yes, this would be unusual because 46% is 6% higher than 40%. c) No, this would not be unusual because the error is only 6%. d) No, this would not be unusual because 46% is only 1.2 standard errors from 40%.

Answer 1

d) No, this would not be unusual because 46% is only 1.2 standard errors from 40%.

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If the z-score is higher than 2 or lower than -2, X is unusual.

In this question:

Mean = 40%. So
`\mu = 0.4`

Standard error = 5%. So
`\sigma = 0.05`

Is 46% unusual?

We have to find Z when X = 0.46. So

`Z = (X - \mu)/(\sigma)`

`Z = (0.46 - 0.4)/(0.05)`

`Z = 1.2`

1.2 is lower than 2, that is, it is only 1.2 standard deviations from the mean. So 46% is not unusual.

So the correct answer is:

d) No, this would not be unusual because 46% is only 1.2 standard errors from 40%.