Question

In a simple random sample of 51 community college statistics students, the mean number of college credits completed was x=50.2 with standard deviation s = 8.3. Construct a 98% confidence interval for the mean number of college credits completed by community college statistics students.

Verify that conditions have been satisfied:

Simple random sample?                            n > 30?

Find the appropriate critical value

Since σ is unknown, use the student t distribution to find the critical value tα2 on table A-3

Degrees of freedom = n – 1 for row

Find α2 for column heading

Locate area

Compute the margin of area E=tα2 sn

Construct the interval   x-E<μ<x+E

Interpret the result.

If σ is known and it is a normally distributed population   OR if σ is known and n > 30

Use the normal (z) distribution.

If the population is not normally distributed and n≤30, use a nonparametric method or bootstrapping method.

Answer 1

Here, the sample size is 51 > 30

So, the population might be assumed to be approximately normally distributed.

The degress of freedom = 51-1 = 50

The confidence level = 0.98

Since, the population standard deviation is unknown, t-distribution should be used.

The critical value = = = 2.403

The margin of error = , where, s = 8.3, n = 51

Hence, margin of error = 2.7928

Thus the 98% confidence interval of mean number of college credits is = [, ] = [50.2 - 2.7928, 50.2 + 2.7928] = [47.4072, 52.9928]