Question

Suppose that the population of the scores of all high school seniors that took the SAT-M (SAT Math) test this year follows a Normal Distribution with mean µ and standard deviation σ = 100. You read a report that says, “On the basis of a simple random sample of 100 high school seniors that took the SAT-M test this year, a confidence interval for µ is 512.00 ± 19.6.” If this is true, then the confidence level for this interval is: a. 90% b. 95% c. 99% d. over 99.9%

Answer 1

Solution:

Given ,

n = 100

σ = 100

also given a confidence interval for  µ is 512.00 ± 19.6

Confidence interval for  µ is given by sample mean ± margin of error

So , we can write ,

margin of error = 19.6

But, margin of error =  /2 * ( / n )

19.6 =   /2 * ( / n )

19.6 =   /2 * (100 / 100 )

  /2 = 1.96

Use z table. See the column of z .See 1.9 and then see the corresponding value at .06

It is 0.975

This is value of 1 - (/2)

1 - (/2) = 0.975

(/2) = 0.025

= 0.05

Now , Confidence level = 1 - = 1 - 0.05 = 0.95 = 95%

Answer : 95%