A wine manufacturer is concerned about the percentage of alcohol in each bottle.A random sample of 31 bottles is tested.The average alcohol percentage is 12.58% and the standard deviation is 0.25%.If wine content is normally distributed,the 90% confidence interval for the average wine content is _____.
A)[12.46,12.70]
B)[12.48,12.68]
C)[12.49,12.67]
D)[12.50,12.66]
E)[12.52,12.64]

Question

Quizplus · Accepted Answer

The 90% confidence interval is calculated using the formula: mean ± (critical value * standard deviation/sqrt(sample size)). With a sample size of 31, the t-distribution critical value for 90% confidence is approximately 1.645. The interval is 12.58 ± (1.645 * 0.25/√31), which results in [12.50, 12.66].

A Wine Manufacturer Is Concerned About the Percentage of Alcohol