Which of the following best represents a 99 percent confidence interval if the mean score from 40 students in an exam is 85 and the population's standard deviation is 18?
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Explanation:
The mean is 85, the standard deviation is 18, and the sample size is 40.
The 99% z-value is 2.58.
Confidence interval = xΛΒ±zβ(nβΟβ)
= 85Β±2.58(40β18β)=[77.658;92.342]
Calculation breakdown:
Standard error = 40β18β=6.324618β=2.846
Margin of error = 2.58Γ2.846=7.342
Lower bound = 85β7.342=77.658
Upper bound = 85+7.342=92.342
Therefore, the 99% confidence interval is [77.658; 92.342].*