1. The 2006 General Social Survey contains information on the number of hours worked by a respondent each week. The mean number of hours worked per week is 39.04, with a standard deviation of 11.51. The sample size is 83. a. Estimate the value of the standard error. b. Calculate a 95 percent confidence interval for the mean number of hours worked per week with these data. c. Say we increased the sample size to 10,000 and found the same mean and standard deviation. What would be the standard error

A)σ_M = 1.2634

B) CI = (36.564, 41.516)

C) σ_M = 0.1151

Step-by-step explanation:

A) Formula for standard error is;

σ_M = σ/√n

We are given;

Mean; x¯ = 39.04

Standard deviation; σ = 11.51

Sample size; n = 83

Thus;

Standard error; σ_M = 11.51/√83

σ_M = 1.2634

B) formula for confidence interval is;

CI = x¯ ± z(σ_M)

z is critical value

At 95% Confidence interval, z = 1.96

Thus;

CI = 39.04 ± 1.96(1.2634)

CI = (39.04 + 2.476) and (39.04 - 2.476)

CI = (36.564, 41.516)

C) sample size is now increased to 10000

Thus,standard error is calculated as;

σ_M = 11.51/√10000

σ_M = 0.1151

could you give more information?

idk the answer but just put a

step-by-step explanation: