Answer:
The answer is false
Step-by-step explanation:
In a sample above 30 obs like this the confidence interval is defined as
X+- t* (s/sqrt(n)) where X is the mean t the tvalue for a given confidence level, n the size of sample and s standar deviation.
To find de appropiate value of t we must see the T table where rows are degrees of freedom and columns significance level
The significance is obtained:
significance = 1 - confidence level = 1 - 0.9 = 0.10
Degrees of freedom (df) for the inteval are
df = n - 1 = 18 - 1 = 17
So we must look for the value of a t with 17 values and significance of 0.10 which in t table is 1.740 not 1.746 ( thats the t for 16 df)
