Question

64 college students took a placement test. Their mean score was 29.5 points with a standard deviation of 7.3 points. Construct a 95% confidence interval for the mean score on this placement exam for all students.

Answer 1

Solution :

The 95% confidence interval for population mean is given as follows :

S Et (0:05_n-1) Vn

Where, x̄ is sample mean, s is sample standard deviation, n is sample size and t_{(0.05/2, n - 1)} is critical t value to construct 95% confidence interval.

We have, n = 64,  x̄ = 29.5, s = 7.3  

Using t-table we get, t_{(0.05/2, 64 - 1)} = 1.998

Hence, the 95% confidence interval for the mean score on this placement exa for all students is,

29.5 + 1.998 x 7.3 64

= (29.5±1.82

= (29.5-1.82, 29.5+1.82)

= (27.68, 31.32)

The 95% confidence interval for the mean score on this placement exa for all students is (27.68, 31.32).