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Answer:
Step-by-step explanation:
Since the sample size is quite large, we can use the z-distribution.
The margin of error is given by
Where n is the sample size, s is the sample standard deviation and is the z-score corresponding to a 90% confidence level.
The z-score corresponding to a 90% confidence level is
Significance level = α = 1 - 0.90= 0.10/2 = 0.05
From the z-table at α = 0.05
z-score = 1.645
Therefore, the margin of error is 0.776.
Answer:
79.5%
Step-by-step explanation: