In: Statistics and Probability
Suppose a math teacher takes a sample of 169 students and calculates their mean math test score to be 78. suppose it is known that the test score is distributed normally with standard deviation of 9.
a. Give 90% confidence interval for u, the population mean, for the average math test score.
b. Interpret the interval in part a. in terms of the problem.
c. what is the value of z for a 98% confidence interval?
d.What is the value of z for 96% confidence interval?
e. How many students must be sampled in order to estimate u within +- 0.2 degree with 90% confidence?
Solution :
Given that,
= 78
= 9
n = 169
At 90% confidence level the z is ,
= 1 - 90% = 1 - 0.90 = 0.10
/ 2 = 0.10 / 2 = 0.05
Z/2 = Z0.05 = 1.645
Margin of error = E = Z/2* ( /n)
= 1.645 * (9/ 169)
= 1.1388
At 90% confidence interval estimate of the population mean is,
- E < < + E
78 - 1.1388 < < 78 + 1.1388
76.8612< < 79.1388
(76.8612 ,79.1388 )
(b)At 98% confidence level the z is ,
= 1 - 98% = 1 - 0.98 = 0.02
/ 2 = 0.02 / 2 = 0.01
Z/2 = Z0.01 =
(c)At 96% confidence level the z is ,
= 1 - 96% = 1 - 0.96 = 0.04
/ 2 = 0.04 / 2 = 0.02
Z/2 = Z0.02 = 2.054