Answer: 2.460
Step-by-step explanation:
The formula of Margin of Error for (n<30):-
[tex]ME=t_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
Given : Sample size : n= 18
Level of confidence = 0.90
Significance level : [tex]\alpha=1-0.90=0.1[/tex]
Using the t-distribution table ,
Critical value : [tex]t_{n-1, \alpha/2}=t_{17,0.05}= 1.7396[/tex]
Standard deviation: [tex]\sigma=\text{ 6 inches }[/tex]
Then, we have
[tex]ME=( 1.7396)\dfrac{6}{\sqrt{18}}\approx2.460[/tex]
Hence, the margin of error (ME) of the confidence interval with a 90% confidence level = 2.460