SOMEONE PLS HELP ME I WILL MAKE U BRAINLIST ! In a survey sample of 83 respondents, about 30.1 percent of the samplework less than 40 hours per week. What is the estimated standard error for the group of respondents who work 40 hours or more per week? (*round to two decimal places)

Answer = √(0.301 × 0.699 / 83) ≈ 0.050

A 68 percent confidence interval for the proportion of persons who work less than 40 hours per week is (0.251, 0.351), or equivalently (25.1%, 35.1%)

Step-by-step explanation:

√(0.301 × 0.699 / 83) ≈ 0.050

We have a large sample size of n = 83 respondents. Let p be the true proportion of persons who work less than 40 hours per week. A point estimate of p is  because about 30.1 percent of the sample work less than 40 hours per week. We can estimate the standard deviation of  as . A  confidence interval is given by , then, a 68% confidence interval is , i.e., , i.e., (0.251, 0.351).  is the value that satisfies that there is an area of 0.16 above this and under the standard normal curve.The standard error for a proportion is √(pq/n), where q=1−p.

Hope this answer helps you :)

Have a great day

Mark brainliest

step-by-step explanation: