Question

Scores on an aptitude test are known to follow a normal distribution with a standard deviation of 32.4 points. A random sample of 12 test scores had a mean score of 189.7 points. Based on the sample results, a confidence interval for the population mean is found extending from 171.4 to 208 points. Find the confidence level of this interval.

Margin of Error (ME)= ?

Z-Score (Z-a/2)= ?

Confidence Level= ?

Answer 1

Solution :

Given that,

Lower confidence interval = 171.4

Upper confidence interval = 208

n = 12

  = (Lower confidence interval + Upper confidence interval ) / 2

= (171.4 + 208) / 2

= 379.4 / 2 = 189.7

= 189.7

Margin of error = E = Upper confidence interval -   = 208 - 189.7 = 18.3

Margin of error = 18.3

sample size = n = (Z_/2* / E) ²

Z_/2 = E * n / = 18.3 * 12 / 32.4 = 1.96

Z_/2 = 1.96

_{= 0.05}

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