You haven't been getting much sleep lately and want to see how you compare to the average CSU student. You plan to sample students and ask them how many hours they sleep per night. Then you will use that data to estimate the population average nightly hours of sleep with an approximate 95% confidence interval. (Critical value 2.) Based on your own sleep, you estimate the standard deviation of hours slept to be 2 hours. What is the minimum number of students you need to sample in order to have a margin of error at most 0.4 hours?

Given Information:

confidence level = 95%

standard deviation = σ = 2

Margin of error = 0.4

Required Information:

Number of students = n = ?

Answer:

Number of students = n ≤ 96 (for z-score = 1.96)

Number of students = n ≤ 100 (for z-score = 2)

Step-by-step explanation:

We know that margin of error is given by

margin of error = z * (σ/√n)

since we want to have a margin of error at most 0.4 hours

margin of error ≤ z * (σ/√n)

√n ≤ z*σ/margin of error

n ≤ (z*σ/Margin of error) ²

Where z is the corresponding z-score from the z-table, n is the number of students required and σ is the standard deviation.

For 95% confidence level the exact z-score is 1.96

n ≤ (1.96*2/0.4) ²

n ≤ 96.04

n ≤ 96

Therefore, a sample size of less or equal to 96 students is required to ensure a margin of error at most 0.4 hours for 95% confidence interval.

If we use the approximate z-score of 2 as given in the question than the required sample size will be

n ≤ (2*2/0.4) ²

n ≤ 100

Therefore, a sample size of less or equal to 100 students is required to ensure a margin of error at most 0.4 hours for 95% confidence interval.