Question

Determine the minimum sample size required to be 95% confident that the sample mean waking time is within 10 minutes of the population mean waking time. Use the population standard deviation from Exercise 1.

Answer 1

Answer:

Step-by-step explanation:

Hello!

Text from Exercise 1:

"The waking times (in minutes past 5:00 a.m.) of 40 people who start work at 8:00 a.m. are shown in the table at the left. Assume the population standard deviation is 45 minutes."

The study variable is X: waking time (past 5:00 a.m.) of one person, measured in minutes.

Assuming the variable has a normal distribution, the margin of error, d, of the CI for the population mean waking time is:

d= $Z_{1-\alpha /2}$ * (δ/√n)

You have to calculate the sample size for a 95% CI, σ= 45min and d=10 min.

For this you have to clear the sample size from the formula:

d= $Z_{1-\alpha /2}$ * (δ/√n)

$\frac{d}{Z_{1-\alpha /2}}$=δ/√n

√n*$\frac{d}{Z_{1-\alpha /2}}$= δ

√n= δ * $\frac{Z_{1-\alpha /2}}{d}$

n= ( δ * $\frac{Z_{1-\alpha /2}}{d}$)²  

n= (45* $\frac{1.96}{10}$)²

n= 77.79≅ 78

You need a sample of at least 78 people to estimate the population mean of the waking time using a 95% CI.

I hope this helps!