You want to estimate the average GMAT score for all students who took the Logitano GMAT Preparation course during the past 2 years.You select a random sample of 12 such students from a comprehensive list of all Logitano students who took the course over the last two years and find that the average GMAT score for the sample was 1040.The population standard deviation is unknown, but the sample standard deviation is 83 points.Assume that the population distribution is approximately normal.Produce the 95% confidence interval estimate of the mean GMAT score for the population of Logitano students over the past two years.Report the upper bound for your interval.
A)925.3
B)988.21
C)1092.74
D)1156.23
E)1193.47

Question

Quizplus · Accepted Answer

To calculate the 95% confidence interval for the mean when the population standard deviation is unknown and the sample size is small (n < 30), we use the t-distribution. The formula for the confidence interval is given by:$$ \bar{x} \pm t_{\alpha/2} \times \left(\frac{s}{\sqrt{n}}\right) $$Where:- $ \bar{x} $ is the sample mean (1040),- $ t_{\alpha/2} $ is the t-score for the 95% confidence level and degrees of freedom (df) = n - 1 = 11,- $ s $ is the sample standard deviation (83),- $ n $ is the sample size (12).Using a t-table or calculator, the t-score for 95% confidence and 11 degrees of freedom is approximately 2.201.The margin of error (ME) is:$$ ME = t_{\alpha/2} \times \left(\frac{s}{\sqrt{n}}\right) = 2.201 \times \left(\frac{83}{\sqrt{12}}\right) \approx 52.74 $$Thus, the confidence interval is:$$ 1040 \pm 52.74 $$The upper bound is:$$ 1040 + 52.74 = 1092.74 $$

You Want to Estimate the Average GMAT Score for All