Question

Here are the summary statistics for ankle girth measurements (in centimeters) for a random sample of...

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", "1.96", "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:

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"]      

Homework Answers

Answer #1

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

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Does anyone still believe the earth is flat? To answer this, a random sample of 55...
Does anyone still believe the earth is flat? To answer this, a random sample of 55 U.S. adults were asked if they believe the earth is flat. Based on the results of the survey, a 95% confidence interval is (0.124, 0.348). (a) What is the estimate of the proportion of all U.S. adults that believe the earth is flat?   (b) Interpret the confidence interval: We are 95% confident the proportion of all U.S. adults that believe the earth is flat...
The heights of a random sample of 50 college students showed a mean of 174.5 centimeters...
The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centimeters. Construct a 98% confidence interval for: a) the mean height b) the standard deviation of heights of all college students. State assumptions.
The heights of a random sample of 50 college students showed a mean of 174.5 centimeters...
The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centimeters. Construct a 98% confidence interval for: a) the mean height b) the standard deviation of heights of all college students. State assumptions.
Use the following information to answer the next three questions. A high school statistics class wants...
Use the following information to answer the next three questions. A high school statistics class wants to estimate the average cookie weight of a generic brand of chocolate chip cookies. They collect a random sample of 50 cookies from the manufacturing process and obtain the weight (in grams) for each cookie. Based on their data, the 95% confidence interval for the average weight per cookie is 25.65 to 26.35 grams. Indicate whether the following is a valid or invalid conclusion....
Pinworm: In a random sample of 830 adults in the U.S.A., it was found that 70...
Pinworm: In a random sample of 830 adults in the U.S.A., it was found that 70 of those had a pinworm infestation. You want to find the 90% confidence interval for the proportion of all U.S. adults with pinworm. (a) What is the point estimate for the proportion of all U.S. adults with pinworm? (b) What is the critical value of z (denoted zα/2) for a 90% confidence interval? zα/2 =   (c) What is the margin of error (E) for...
Pinworm: In a random sample of 830 adults in the U.S.A., it was found that 74...
Pinworm: In a random sample of 830 adults in the U.S.A., it was found that 74 of those had a pinworm infestation. You want to find the 90% confidence interval for the proportion of all U.S. adults with pinworm. (a) What is the point estimate for the proportion of all U.S. adults with pinworm? Round your answer to 3 decimal places. (b) Construct the 90% confidence interval for the proportion of all U.S. adults with pinworm. Round your answers to...
Here are summary statistics for randomly selected weights of newborn girls: n = 150 , x...
Here are summary statistics for randomly selected weights of newborn girls: n = 150 , x = 30.5 hg, s = 6.7 hg. Construct a confidence interval estimate of the mean. Use a 90% confidence level. Are these results very different from the confidence interval 29.4hg < μ < 31.4 hg with only 19 sample values, x = 30.4 hg, and s = 2.5 hg? What is the confidence interval for the population mean μ? ___________hg < μ < __________...
Here are summary statistics for randomly selected weights of newborn​ girls: n equals 202​, x overbar...
Here are summary statistics for randomly selected weights of newborn​ girls: n equals 202​, x overbar x = 33.7 ​hg, s = 7.3 hg. Construct a confidence interval estimate of the mean. Use a 90​% confidence level. Are these results very different from the confidence interval 31.8 hg < mu < 34.6 hg with only 17 sample​ values, x overbar x = 33.2 ​hg, and s = 3.2 ​hg? What is the confidence interval for the population mean mu μ​?...
The accompanying summary data on total cholesterol level (mmol/l) was obtained from a sample of Asian...
The accompanying summary data on total cholesterol level (mmol/l) was obtained from a sample of Asian postmenopausal women who were vegans and another sample of such women who were omnivores. Diet Sample Size Sample Mean Sample SD Vegan 89 5.20 1.08 Omnivore 97 5.65 1.10 Calculate a 99% CI for the difference between population mean total cholesterol level for vegans and population mean total cholesterol level for omnivores. (Use μVegan − μOmnivore. Round your answers to three decimal places.) ,...
Here are summary statistics for randomly selected weights of newborn​ girls: n equals=171​, x over barx...
Here are summary statistics for randomly selected weights of newborn​ girls: n equals=171​, x over barx equals=30.6 hg, s equals=6.56.5 hg. Construct a confidence interval estimate of the mean. Use a 90​% confidence level. Are these results very different from the confidence interval 29.629.6 h gless than<muμless than<31.231.2 hg with only 20 sample​ values, x overbarx equals=30.4 hg, and s equals=2.12.1 hg? What is the confidence interval for the population mean hg ​(Round to one decimal place as​ needed.)