Question

Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation σ . Assume that the population has a normal distribution. College students' annual earnings:
98% confidence;
n = 9,
mean = $3211,
s = $897

Answer 1

Answer:

College students' annual earnings is betweeen $2,514.33 and $3,907.67

Step-by-step explanation:

Hi, first, let´s introduce the formula that we need to use in order to find out the lower and higher limit of the interval.

$HigherLimit=Average+Z_{\frac{\alpha }{2} }\frac{Sigma}{\sqrt{n} }$

$LowerLimit=Average-Z_{\frac{\alpha }{2} }\frac{Sigma}{\sqrt{n} }$

Now, the only problem here is to find Z(alpha/2), so, let´s define that. Alpha is the remeining area of the normal distribution curve that is out of the limit, in other words, in this case, we need the 98% confidence interval, that means that 2% of the probs are going to be out of the range, therefore, this 2% is alpha. Since we need that the same area is removed from the left and right side of the uniform distribution curve, we need to find the value of Z in the uniform curve distribution table for 1% (that is 2%/2) and due to symmetry, we can now find the values of the interval (this means that we need the value of Z for 0.99, that is 2.33). Now, let´s find out the lower and higher limit of this interval.

$HigherLimit=3,211+(2.33)\frac{897}{\sqrt{9} }=3,907.67$

$LowerLimit=3,211-(2.33)\frac{897}{\sqrt{9} }=2,514.33$

So, the college students´annual earnings are between $2,514.33 and $3,907.67 with 98% of confidence.