Question

Here are the summary statistics for ankle girth measurements (in centimeters) for a random sample of 50 U.S. adults. Summary statistics: Column Mean Std. dev. n Ankle_girth 22.16 cm 1.86 cm 50 What is the estimated standard error for sample means from samples of this size? (rounded to two decimal places) What is the critical T-score for a 90% confidence interval? Inverse T-distribution Calculator We used StatCrunch to find the 90% confidence interval. Mark each interpretation of this confidence interval as valid or invalid. One sample T summary confidence interval: 90% confidence interval results: Mean Sample Mean Std. Err. DF L. Limit U. Limit μ 22.16 0.26 49 21.7 22.6 We are 90% confident that the mean ankle girth for the population of all U.S. adults is between 21.7 and 22.6 centimeters. We are 90% confident that the mean ankle girth for this sample of 50 U.S. adults is between 21.7 and 22.6 centimeters. We should not estimate the mean ankle girth using StatCrunch because the conditions are not met for use of the T-model to represent the distribution of sample means.

Answer 1

Answer:

Step-by-step explanation:

1) to find the estimated standard error for sample means from the size sample

= 1.86/sqrt(50)

= 0.26

2) to evaluate the critical T-score for 90% confidence interval = 1.6766

3) we have 90% confidence that the mean ankle birth for the population of all U.S adults are between the range of 21.7 and 22.6 centimeters.