Question

Suppose there is a population of test scores on a large, standardized exam
for which the mean and standard deviation are unknown. Two different random
samples of 50 data values are taken from the population. One sample has a
larger sample standard deviation (SD) than the other. Each of the samples is
used to construct a 95% confidence interval. How do you think these two
confidence intervals would compare?

Answer 1

Higher the standard deviation results into higher value of Margin of error, ME
ME = z*s/sqrt(n)
as z and n are same for both samples, higher value of s will result into higher value of ME.

This means that the sample with higher standard deviation will result into wider confidence interval considered to the other sample.

Answer 2

ANSWER :

At 95% confidence level,  interval is z = - 1.96 to z = + 1.96. For both samples this z - value limits will be same. 

Now z = (x - m) / SD 

=> x = z * SD + m 

So, x limits will be wider for higher SD compared to that with lower SD at the given confidence level (95% in this case).