Question

QUESTION 3 For this question we will analyze data from the most detailed, up-to-date portrait of ho poiple in the United States spend their time- it American Time Use Survey. The survey offerns also offers some insight into how Amexica ns veicise and eat. I have included all the information you need to answer this question without yoar nooding to access the data yourself (although it is publicly available). The data are drawn from a sample of n- 11,592 but wo know that the population of the United States is something like N 320 million. 1. What assumption is required in order for us to be able to draw inferencss about the 320 million people living in the United States from the 11,592 individuals in this sample? The ATUS asks respondents to self-report the total time (in minutes) they spend in primary eating and drinking as well as the total time they spend snacking. The sum of these two variables constitutes the total eating cime in the data. The mean and standard deviation of the TOTAL EAT variable are = 82.56 and s = 66.75 minutes. 2. What is the standard error of the TOTAL EAT variable? Explain what the standard errotr means in words. 3. Does the standard error depend on the population size, N7 4. Set a 0.05-that is, you are willing to be wrong 5% of the time. (a) Construct a 95% confidence interval for TOTAL-ET. (b) Say what this interval tells you in words. 5. Now set ? 001. (a) Construct a 99% confidence interval for TOTAL-EAT. (b) Say what this interval tells you in words.
#4 please

Answer 1

4)

The provided sample mean is X = 82.56 and the sample standard deviation is s = 66.75. The size of the sample is n = 11592 and the required confidence level is 95%. The number of degrees of freedom are df- 11592-1 11591, and the significance level is -0.05 Based on the provided information, the critical t-value for α = 0.05 and df = 11591 degrees of freedom is te 1.96 The 95% confidence for the population mean μ is computed using the following expression te XS Vn Therefore, based on the information provided, the 95 % confidence for the population mean Al is 쀼쁑82.56 1.96 x 66.75 1.96 x 66.75 cr = (82.56 ) - (82.56-1.215, 82.56+1.215) (81.315,83.)

with 95% confidence we can say that the population mean lies between 81.345 to 83.775