A sample of = 21 of juniors at PSU found a mean age of 23 years

Question

A sample of n1 = 21 of juniors at PSU found a mean age of 23 years. A sample of n2 = 20 juniors at OSU found a mean age of 22 years. The population ages are normally distributed and o1 = 2 o2 = 5.

Find a 95% confidence interval in the difference in the mean age, round answers to to 4 decimal places.

< u1-u2 <

Answer 1

Since we're assuming the two distributions are normally distributed with known σ² values, we'll use a normal critical value of 1.96 instead of a critical value from a t-table. Then the formula for our confidence interval is as follows:

μ₁ - μ₂ ± z_α/2•√(σ₁²/n₁+σ₂²/n₂)

So plugging in all those values we get:

23-22±1.96•√(4/21+25/20)

1±1.96•1.2002

1±2.3524

(-1.3524, 3.3524)

So we can be 95% confident that the mean difference in ages is between -1.3524 and 3.3524. If you're needing something about the two colleges being significantly different in ages, note that zero is included in this interval.