Suppose you have a random sample of 200 students' GMAT scores that have a sample mean of 700 and sample standard deviation of 50, and would like to calculate the 90 percent confidence interval of the population mean. Which of the following would be the correct construction?
A) 700 ± 1.65 $\left( \frac { 50 } { \sqrt { 200 } } \right)$
B) 700 ± 1.96 $\left( \frac { 50 } { \sqrt { 200 } } \right)$
C) 700 ± 1.65 $\left( \frac { 50 } { 200 } \right)$
D) 700 ± 1.96 $\left( \frac { 50 } { 200 } \right)$

Question

Quizplus · Accepted Answer

The correct construction for a 90% confidence interval involves using a z-score (since the sample size is large) that corresponds to the desired level of confidence. For a 90% confidence interval, the z-score is approximately 1.65. The formula for the confidence interval is the sample mean ± z * (sample standard deviation / sqrt(sample size)). Therefore, option A is correct because it uses the 1.65 z-score and correctly applies the formula for the standard error of the mean, which is the sample standard deviation divided by the square root of the sample size.

Suppose You Have a Random Sample of 200 Students' GMAT (50200)\left( \frac { 50 } { \sqrt { 200 } } \right)(200​50​)

Suppose You Have a Random Sample of 200 Students' GMAT $\left( \frac { 50 } { \sqrt { 200 } } \right)$