Answer: 7.1% and 6%.
Explanation:
Given data:
Sample size (n) = 11
Standard deviation = 2.2
Margin of error confidence level 95% = 1.96
Margin of error confidence level of 90% = 1.65
Solution:
P = 2.2/11
= 0.2.
ME = z √p (1-p)/n
at 95% interval
ME = 1.96 * √ 0.2 ( 1 – 0.2 ) / 11
= 1.96 * 0.03636
= 0.071.
= 7.1%
At 90% interval
ME = 1.65 * √ 0.2 ( 1 – 0.2 ) / 11
= 1.65 * 0.03636
= 0.06.
= 6%