Answer: The margin of error for a 95% confidence interval around this sample mean is 1.412.
Step-by-step explanation:
We know that the formula to find the margin of error is given by :-
[tex]E=\pm z^*{\dfrac{\sigma}{\sqrt{n}}}[/tex], where [tex]\sigma[/tex] = population standard deviation
n= sample size , z* = critical z-value as per confidence level.
As per given , we have
[tex]\sigma[/tex] =4.2
n= 34
By z-table , for 95% confidence level : z* = 1.96
Then, the margin of error will be :
[tex]E=\pm (1.96)\dfrac{4.2}{\sqrt{34}}\approx\pm1.412[/tex]
Hence, the margin of error for a 95% confidence interval around this sample mean is 1.412.
