Question

a sample of 66 college students yields a mean annual income of $4035. assuming that the population standard deviation is $875, find a 99% confidence interval for the mean annual income of all college students and interpret the interval.

Answer 1

The 99% confidence interval for the mean annual income of all college students is ( 3,757.55, 4,312.45).

The confidence interval formula is expressed as:

Confidence interval = x ± Z( σ / √n )

Given the parameter:

Sample size n = 66

Sample mean x = $4035

Standard deviation σ = $875

Note that, the Z-score for a 99% confidence level is approximately 2.576.

Hence Z = 2.576

Now, plug these values into the above formula and simplify:

Confidence interval = x ± Z( σ / √n )

Confidence interval = 4035 ± 2.576( 875 / √66 )

Confidence interval = 4035 ± 2.576( 107.705 )

Confidence interval = 4035 ± 277.45

Hence:

Lower Bound = 3,757.55

Lower Bound = 4,312.45

Therefore, we are 99% confident that the true mean annual income of all college students is within the calculated interval ( 3,757.55, 4,312.45).

There is a 1% chance that the true mean annual income is less than $3,757.55 or greater than $4,312.45.