Question

The average hourly wage for a high school student in Illinois is $8.25 per hour with a standard deviation of $.25. Assume hourly wages vary normally.
A) Determine the margin of error for this scenario.
B) For the previous question you just answered, construct a 95% confidence interval for the hourly wage for a high school student.
C) For the 95% confidence interval you just constructed, does an hourly wage of $9.15 fall within that acceptable range?

Answer 1

Answer:

Step-by-step explanation:

Hello!

The variable of interest is

X: hourly wage of a high school student

X~N(μ;σ²)

X[bar]= $8.25

σ= $0.25

For a 95% CI for the average hourly wage for 1 high school student:

For this exercise you have to consider the sample size n=1

a) The margin of error of the CI is calculated as:

d= $Z_{1-\alpha /2}$ * σ

1-α: 0.95 ⇒ α:0.05

1-α/2:0.975

$Z_{1-\alpha /2}= Z_{0.975}= 1.96$

d= 1.96 * 0.25= 0.49

b)

The formula for the interval is

X[bar] ± $Z_{1-\alpha /2}$ * σ

[8.25 ± 0.49]

[7.76; 8.74]

With a 95% you'd expect the interval $[7.76; 8.74] to include the average hourly wage of one high school student in Illinois.

c)

Te calculated interval is $ [7.76; 8.74] the value $9.15 does nor fall within that acceptable range.

I hope this helps!