Given Information:
Standard deviation = σ = $1000
Mean = μ = $10,979
Confidence level = 90%
Margin of error = $250
Sample size = n = 20
Required Information:
a) 90% confidence interval = ?
b) Sample size at 99% confidence = n = ?
Answer:
a) 90% confidence interval = ($10,611.17, $11,346.83)
b) Sample size at 99% confidence = n = 107
Step-by-step explanation:
a) The confidence interval is given by
confidence interval = μ ± z*σ/√n
Where μ is the mean annual health insurance premium, σ is the standard deviation, n is number of subscribers that were surveyed, and z is the corresponding z-score for 90% confidence level.
The z-score for 90% confidence level is 1.645
Confidence interval = 10,979 ± (1.645*1000)/√20
Confidence interval = 10,979 ± 367.83
upper limit = 10,979 + 367.83 = 11,346.83
lower limit = 10,979 - 367.83 = 10,611.17
Confidence interval = ($10,611.17, $11,346.83)
Which means that we are 90% confident that the annual health insurance premiums for a family with coverage through an employer is between ($10,611.17, $11,346.83)
b) To find the required sample size
Me = z*(σ/√n)
√n = z*σ/Me
n = (z*σ/Me)²
Where z is the corresponding z-score, n is the required sample size σ is the standard deviation and Me is the margin of error that is $250
The z-score for 99% confidence level is 2.576
n = (2.576*1000/250)²
n = 106.17
n = 107
Therefore, a sample size of 107 is needed to find the population mean within $250 at 99% confidence.
or ≈ $562.43