In a survey sample of 83 respondents, about 30.1 percent of the sample work less than 40 hours per week. Calculate a 68 percent confidence interval for the proportion of persons who work 40 hours or more per week.

The 68% confidence interval would be given by (0.370;0.470)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 68% of confidence, our significance level would be given by and . And the critical value would be given by:

The proportion os persons who work 40 hours or more is 1-0.301= 0.699

The confidence interval for the mean is given by the following formula:  

If we replace the values obtained we got:

The 68% confidence interval would be given by (0.370;0.470)

answer: it is 1: 47 pm

the answer is the option b

the range is all real numbers

step-by-step explanation:

we know that

the absolute value parent function is equal to

see the attached figure to better understand the problem

so

a. it is v-shaped > is true

b. the range is all real numbers > is not true

because the range is the > [0,∞)

all real numbers greater than or equal to zero

c. it goes through the origin > is true

d. the domain is all real numbers > is true

step-by-step explanation: