Assuming the population has an approximate normal distribution, if a sample size n = 17 has a sample mean x = 47 with a sample standard deviation s = 3 , find the margin of error at a 98 % confidence level. Round the answer to two decimal places.
Please help
You can play around with Z table stuff at
http://davidmlane.com/hyperstat/z_table.html
To find the margin of error at a 98% confidence level, we need to use the formula:
Margin of error = Z * (s / sqrt(n))
Where:
Z is the z-score associated with the desired confidence level,
s is the sample standard deviation,
and n is the sample size.
First, let's find the z-score for a 98% confidence level. The z-score can be obtained from a standard normal distribution table or calculated using software. For a 98% confidence level, the z-score is approximately 2.33.
Next, we can substitute the given values into the formula:
Margin of error = 2.33 * (3 / sqrt(17))
Now, we need to calculate the square root of 17:
sqrt(17) ≈ 4.12
Substituting this value into the formula:
Margin of error ≈ 2.33 * (3 / 4.12)
Simplifying the expression:
Margin of error ≈ 2.33 * 0.73
Calculating the product:
Margin of error ≈ 1.7
Rounding this value to two decimal places, the margin of error is approximately 1.70.
So, at a 98% confidence level, the margin of error is approximately 1.70.