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)                            [ Select ]                       ["1.86", "0.26", "0.04"]      

What is the critical T-score for a 90% confidence interval?                            [ Select ]                       ["1.645", "1.6759", "1.6766"]         
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.                            [ Select ]                       ["valid", "invalid"]      

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.                            [ Select ]                       ["valid", "invalid"]      

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.                            [ Select ]                       ["valid", "invalid"]      

Answer 1

1) estimated standard error for sample means from samples of this size = 1.86/sqrt(50)=0.26

2) critical T-score for a 90% confidence interval =1.6766

3) 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. valid