Question

A random sample of 8 recent college graduates found that starting salaries for architects in New York City had a mean of $42,653 and a standard deviation of $9,114. There are no outliers in the sample data set. Construct a 95% confidence interval for the average starting salary of all architects in the city.
A. (35222.41, 50083.59)
B. (34506.12, 50799.88)
C. (36337.32, 48968.68)
D. (35032.29, 50273.71)

Answer 1

Answer:

C. (36337.32, 48968.68)

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.025 = 0.975$, so $z = 1.96$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{9114}{\sqrt{8}} = 6315.68$

The lower end of the interval is the sample mean subtracted by M. So it is 42653 - 6315.68 = 36337.32.

The upper end of the interval is the sample mean added to M. So it is 42653 + 6315.68 = 48968.68.

So the correct answer is:

C. (36337.32, 48968.68)