Here are the summary statistics for ankle girth measurements (in centimeters) for a random sample of 50 U.S. adults.
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", "1.96", "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:
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"]
1)
a) What is the estimated standard error for sample means from samples of this size=std dev/√n
=1.86/sqrt(50)=0.26
b) critical T-score for a 90% confidence interval =1.6766
2)
a)
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. : Vaild
b)
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. :Invalid (interval is for population mean and not sample mean)
c)
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. : Invalid
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