A human resource manager for a large company takes a random sample of 60 employees from the company database. Based on the sample she calculates a 95% confidence interval for the mean time of employment for all employees to be 8.7 to 15.2 years. Which of the following will provide a more informative (i. e., narrower) confidence interval than the 95% confidence interval? A. Using a 90% confidence level (instead of 95%)
B. Using a 99% confidence level (instead of 95%)
C. Using a sample size of 40 employees (instead of 60)
D. Using a sample size of 90 employees (instead of 60)

A. Using a 90% confidence level (instead of 95%)

D. Using a sample size of 90 employees (instead of 60)

Step-by-step explanation:

The margin of error of a confidence interval is given by:

In which z is related to the confidence level, is the standard deviation of the population and n is the size of the sample.

The higher the margin of error, the less precise the confidence interval is.

We have:

A 95% confidence interval, with a sample of 60.

We want to make it more precise:

Two options, decrease z(decrease the confidence level), or increase n(increase the sample size).

So the correct options are:

A. Using a 90% confidence level (instead of 95%)

D. Using a sample size of 90 employees (instead of 60)

tangent

step-by-step explanation:

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116